Oxford Admissions Question (No Calculator)

Can you estimate the log of 3 to the base of 2 without using a calculator? This kind of skill is actually pretty important in university mathematics. Watch the video to learn how to solve it!
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[@castravetdragos4012]

So ... I have a 25% chance to enter the Oxford .... *Cool*

[@10monkeybutt]

Beats a 1/1,024th of a chance.

[@shreyartz1697]

I think you are good at probability in mathematics not logarithms 😊😄😄

[@AagneyWest]

Getting correct ans for 1 question is 1/4 but for 2 question itself it becomes 1/16, guess how many questions will be there in test..then prob will be 1/2to the power x.

[@lordwalterebenyzrbautista204]

@@AagneyWest You mean 1/4^x?

[@AagneyWest]

@@lordwalterebenyzrbautista204 yeah

[@mindlessintellectual2097]

I thought we were going to calculate the actual result

[@garethb1961]

I tried to do that for a half hour after looking at the video thumbnail without watching the video. Always read the problem, eh?

[@joshscores3360]

@@garethb1961 ≈ ≠ =

[@henrychan720]

You can do a Taylor expansion and substitute

[@NoActuallyGo-KCUF-Yourself]

The actual value IS log_2(3). Any decimal representation is only an approximation of an irrational number.

[@brandaccount8293]

Well I was able to do in 30 seconds (30 seconds just to divide value of log2 by log3 (base 10) )

[@JohnnyReb1976]

"Morty, what is 5 times 9?" "It's ummm at least forty."

[@JohnnyReb1976]

@Julia Li Rick and Morty s1 e4

[@plantman1093]

I know it's less than 50

[@mt9456]

@@plantman1093 I will use my genius intellect to calculate 5 times 9. if its bigger than forty and smaller than 50... it should be... in between. or 45 to be exact

[@plantman1093]

@Țŕicķster nah he just got lucky

[@ViratKohli-jj3wj]

Gravity Falls is better than rick and morty

[@Korbinho]

Nobody: KZbin Recommendations: Aight, it’s time for you to apply for Oxford..

[@ijemand5672]

Next time get rid of the useless nobody, maybe someone will laugh

[@flclub54]

^ true

[@Dankschon]

@@ijemand5672 U mad?

[@zordican3916]

Aigt Ima make this man's whole career.

[@sonicalibur_iii]

What? No! Keep the "Nobody"

[@ar00042]

What’s 2+2 (no calculator) *Starts sweating😨😰

[@carloszamorano8297]

5 😎

[@deineroehre]

There is a way to square root this and getting some functions applied which then lead to the solution. Just ask presh talwalkar for the most complicated solution, he will deliver!

[@yurenchu]

- "What’s 2+2 (no calculator) " I don't know, but 2(2+2) amounts to 0.5 , according to Presh's "modern" calculator. - kzbin.info/www/bejne/rJLMpaZ4it6chLM

[@Peter_1986]

Obviously 22.

[@PositronQ]

Atharva Raje 3

[@Camronnader1998]

Sad thing is that I used to know how to do that lmao... crazy how fast you can forget something if you don't continuously practice it

[@koro-sensei9783]

I too knew it

[@dnhdfnfkrjxjxfjjggj3002]

Practice makes a man perfect

[@sciuresci1403]

How to do what ?

[@user-ri8ps6cl4w]

@@dnhdfnfkrjxjxfjjggj3002well clearly practicing a month for a test was not enough :(

[@irrelevant_noob]

@@sciuresci1403 solve the problem. (so basically... Apply the technique shown in the video.)

[@Catsrock46060]

I actually tried to apply for Oxford (although I didn't get in) and found questions like these the easiest and rather fun to do. Usually all the problems were rather unique, but they did get rather difficult at the end, wouldn't mind seeing some solutions of those, they really made you think.

[@KasperMeerts]

From music theory, we know that a perfect fifth is almost exactly 7 semitones. A perfect fifth also stands for a frequency ratio of 3/2. So 3/2 has to be very close to 2^(7/12), meaning 3 has to be close to 2^(19/12), hence 2log 3 should be about 19/12 = 1.583...

[@davem3953]

nerd^2 :)

[@ffggddss]

I also thought of that. (And it's the reason we divide the octave into 12 semitones - because 19/12 is the "best" fractional approximation for log₂3, to no better precision than nearby frequencies distinguishable by the human ear.) So that 2¹⁹ = 524288 ≈ 531441 = 3¹² - take the 12th root on both sides, and the ratio gets even closer to unity, as seen by: 2^(19/12) = 2.996614... ≈ 3 Fred

[@magicbean4616]

@@ffggddss doing this kind of approach but looking for smaller powers that were close is how I solved it. Like 2^19 is approximately 3^12 you can find that 2^8 =256 is close to 243=3^5. Using these powers instead gives you an approximation of 1.6. Or the other way 2^(8/5)=3.03...

[@dudewaldo4]

Ooooh spicy

[@ffggddss]

​@@magicbean4616 Yes, that works; and you can streamline the whole process by using continued fractions on the computed value of log₂3 [which does require a calculator!]. In my school days, what you would do here, is to use the common logs you had memorized (usually to just 3 decimals; 5 places for the real nerds). If you didn't have them memorized, you whipped out your slide rule and used the C/D & L scales to look them up: log₁₀2 = .301...; log₁₀3 = .477... Then, 477/3 = 159, so log₂3 = log₁₀3/log₁₀2 ≈ 477/301 ≈ 1.59 or 1.58 That said, it can still be a lot of fun seeing how well you can find these values without a calculator, just by comparing some small integer powers of small integers (and maybe sneaking in a memorized value here or there). Examples: 2¹⁰ = 1024; log₁₀2 ≈ 3.01/10 = .301 [ln(10) ≈ 2.3; so log₁₀(1.023) ≈ ln(1.023)/2.3 ≈ .023/2.3 = .01] 2¹⁶ = 65536; 3⁸ = (((3²)²)²) = ((9²)²) = (81²) = 6561 log₁₀6561 ≈ log₁₀65536 - 1 = 16 log₁₀2 - 1 ≈ 3.816; log₁₀3 ≈ 3.816/8 = .477 7⁴ = ((7²)²) = (49²) = 2401 ≈ 2³·3·10² 4 log₁₀7 ≈ 3(.301) + .477 + 2 = 3.38; log₁₀7 ≈ 3.38/4 = .845 13³ = 2197; 3⁷ = 2187; log₁₀13 ≈ (7/3)log₁₀3 ≈ 7(.159) = 1.113; Correction: 2197/2187 ≈ 1 + 1/220 ≈ 1.0046; log₁₀(1.0046) ≈ ln(1.0046)/2.3 ≈ .0046/2.3 = .002 So log₁₀(2197) ≈ log₁₀(2187) + .002 ≈ 3.339 + .002 = 3.341 log₁₀13 ≈ ⅓(3.341) ≈ 1.114 Fred

[@jerry2357]

This was my engineer’s pragmatic approach: Without doing any calculation, from seeing the thumbnail my answer was “a bit over 1.5”. My reasoning was as follows: log to base 2 of 2 is 1, log to base 2 of 4 is 2, and linear interpolation to get log to base 2 of 3 between these known figures would give 1.5. But I know that the divisions on log graph paper start large and get smaller (knowledge that comes from analysing experimental results before there were computers readily available everywhere), so the actual answer must be larger than the linear interpolation figure. If I was pressed for time in an exam, I would thus guess (b) and move on. Without doing more analysis than my initial reasoning it’s possible that the answer might be above 5/3, i.e. answer (c), but my gut feel was that this figure is a bit high (this was proved by Presh formally in the video). If I needed the answer precisely I would use a calculator or computer.

[@nihalbadiger8922]

It'll be approximately equal to 1.6 You can do it by doing this: (log3 to the base 10)/(log2 to the base 10) Which will be 0.48/0.30 The result will come out to be approximately 1.6 But in the end you will need to know the value of log2 and log3 to the base 10. If you r a science student then you might know those values.

[@jerry2357]

@@nihalbadiger8922 I've been using calculators and computers to do my calculations for over 40 years (although I used log tables up to the age of 16 because calculators weren't allowed in examinations until after 16 in those days), so I don't remember logs these days (other than the obvious ones, whole number powers of the logarithm's base).

[@vasylsky9486]

I did similarly but with better validation of result: 2^(3/2) = 2*Sqrt(2) ~ 2.8. So it has to be a bit higher than 1.5

[@hellobye9178]

That's what I thought too, but for a slightly different reason: 2^1.5 = 2 * √2 √2 ~ 1.414 2^1.5 ~2.828 < 3 But it's close, so I would guess that it's B, not C.

[@hellobye9178]

@@vasylsky9486 Oh woops. I did the same thing as you but replied before looking at the comments. Sorry.

[@kendots6654]

With choices this was very easy. I was having a hard time thinking that they wanted some exact solution (I was solving it based on the thumbnail).

[@sebastianpochert4511]

log_2 of 3 IS the exact solution.

[@trunc8]

Can be done to an approximation. Compare higher powers of 3 and 2 that are close. (all log in base 2) log3=(1/5)*log(243)≈0.2*log(256)=0.2*8=1.6 which is pretty darn close to the answer 1.58

[@SirTylerGolf]

@@trunc8 i just figured out that it was a bit higher than 1.5, guessed 1.6

[@SirMrTreflip]

That 'equals to' symbol indicates that it's not an exact answer

[@Kernel15]

@@trunc8 That's pretty much what Presh is doing.

[@meera1158]

sqrt(2) is 1.4, so double that is 2.8, and 2*sqrt(2) is 2^1.5 so the answer is just slightly above 1.5

[@McRobPoint]

Arsacid That’s exactly how I addressed this problem. I still wasn’t sure if it was b, because c) could have been possible, too (I don’t know by heart what the 3rd root of 4 is). B seemed the by far the better guess, though :-)

[@bengtbengt3850]

sqrt(2) is not 1.4, so your reasoning is not quite right. Alternatively, 2*sqrt(2) = sqrt(4)*sqrt(2) = sqrt(8) < sqrt(9) = 3. 2*sqrt(2) = 2^(3/2), so 2^(3/2) < 3, so 3/2 < log_2(3). Of course, that’s not the end of the solution. You still have to prove the upper bound.

[@michelsfeir1127]

Bengt Bengt "sqrt(2) is not 1.4" it's near enough as makes no matter

[@meera1158]

Bengt Bengt Michel Sfeir Bengt Bengt yeah but its close enough to 1.4, we are estimating. In a larger exam where there are other questions, estimating and using an educated guess (b vs c was the ambiguity in my solution but b wouldve been my guess) on a multiple choice question is by far the quicker way to get to the solution and get to the other questions. if you have time by the end, you can always go back and prove both bounds.

[@vhandarsh51]

That's how I went but I had trouble with upper limit tho

[@Errichto]

Your solution seems to be about guessing that we need to raise to a random power while his can be done easier: log_2(3) < 5/3 (make both sides 2^x) 3 < 2^(5/3) (raise both sides to 3rd power) 3^3 < 2^5, done

[@Martin__]

Yeh, I was so confused about his way of thinking. What you did is what I learned in class.

[@silyu97]

ye it is simple. it doesn't require not only a calculator but also not writing anything on paper. pretty easy mental math

[@thedoctordowho2022]

Can any one explain me this way?

[@magmusacy4937]

How did you come up with this 5/3 ?

[@chandy3859]

​@Ayanokouji Kiyotaka there's 4 choice in the answer. he could try it one by one

[@chessandmathguy]

0:25 is option (d) a joke? It's obviously less than 2 since log 4 with base 2 is 2.

[@MKProsD]

It's the pity option oxford allow you to eliminate for taking the test

[@chessandmathguy]

@@MKProsD yeah they throw you a freebie to help you out. If you can't eliminate (d) immediately, then this test is not for you lol.

[@kienthanhle6230]

If u can't eliminate (d),you're not worth to go to Oxford. Eliminate wrong answer is a basic skill in solving problem

[@helbertrodriguez6449]

Fell pretty good

[@noahali-origamiandmore2050]

Agreed

[@dtrcs9518]

You don't even need to find out the bounds exactly, just by looking at the alternatives you can see it's going to be B

[@mahodaadikari4116]

I also did the same. But finding the bounds is the better way of solving this type of problems.

[@fulltimeslackerii8229]

Wow you’re so cool.

[@bikramadityarajendrapatra3447]

I didn't understand

[@SergioNilo]

yeah, aways is the B... But in case of quantum mechanics questions, u will need to know if it is AB, AC, AD, BC, BD or DC. But when u found the answer it will Just be A, B, C or D again in your universe.

[@michaelnajera901]

@@SergioNilo No te entendi nada

[@zacharymesecke9638]

Every time I see a maths problem on youtube I try to solve it and I'm always really dissapointed in my problem solving capabilities. I just wanna be smart

[@manuelmatias3772]

Did you watch the "lockdown lecture" on problem solving? Searching those terms should return the video I'm talking about.

[@theWebWizrd]

Failing to solve a problem like this doesn't have to mean anything about how smart you are :) It is more probable that it's just something you haven't seen before / been taught. If you keep looking and keep learning you will get there, it just takes time.

[@zacharymesecke9638]

@@manuelmatias3772 I'll check it out, thanks

[@zacharymesecke9638]

@@theWebWizrd thanks mate

[@user-ri8ps6cl4w]

bro you are amazing for trying i dont even try... and plus thats just math you have a thousand other things to be smart with

[@lucyschneider463]

For everybody wondering why this is so easy, even though it's from Oxford: The Oxford admissions process consists of multiple steps, one is a written test. The test is there, to select candidates for the interviews. In this test (called the MAT for maths admissions test) there are two sections, one multiple choice section (they're just easier to mark, given the sheer amount of applicants oxford has, this is really needed) and one section with long answers. This question is one of the many multiple choice questions from the mat, being able to solve this is nice, but that doesn't say anything... (I am an Oxford maths student, if you got questions about it, pls just ask me)

[@th5965]

Lucy Schneider does Oxford care about your GCSE can someone who gets all 6’s but all a*s in A level get in

[@amritlohia8240]

@@th5965 GCSEs are one factor that is taken into account, but they aren't looked at in isolation - they also take into account how good or bad your school was, and how good or bad your GCSE results were compared to the average results within your school. However, by far the most important factors are your MAT score and how well you perform in interview. If you get through those hurdles, and thereby get an offer of a place, then it's pretty much expected that you will get A*s at A-Level, since the MAT and the interview are both more difficult than A-Level.

[@0megazeero]

"easy" Ok boomer

[@SA-wl3ny]

I want be a writer and professor of English graduated from oxford. Will I also have to give this exam ?

[@zulsyahril6428]

I have a question how do you exercise to be more creative in getting creative solutions in math as an oxford student ?

[@Krebzonide]

Just by knowing the square root of 2 I knew it had to be just over 3/2 so I picked the right answer without knowing the upper bound.

[@eganeshseshadri6441]

Did exactly the same way - 2^1.5 should be just over 2.8, so went with b without working out the upper bound.

[@kevinstreeter6943]

@@eganeshseshadri6441 I did it the same way. I put my solution in a comment.

[@godcolor9722]

I don't understand a thing why do we need to take 2^1.5 please help

[@kevinstreeter6943]

@@godcolor9722 Log3 is an exponent when placed over 2 you get 3. SQRT2 = 1.414. This is a good number to memorize. Looking at the lower bound in answer b, 1 1/2, which is 1.5. Place that over 2. 2^(1+1/2). Using the rules of exponents: 2^(1+1/2) = 2*2^(1/2) = 2*sqrt2=2*1.414. Which is slightly less than 2*1.5 = 3. b seems like a reasonable answer. Does that help? I am limited with typing a solution.

[@JaviMoreno5]

@@kevinstreeter6943 But then you cannot discard option c)

[@mjones207]

One advantage to having been in high school before students were allowed to use calculators (I took algebra 2 in 1978) is that certain functions were used often enough that memorizing was faster than looking them up in a table. For example, common log values of log(2) ≈ 0.30103 and log(3) ≈ 0.47712. (Knowing these gave values for common logs of 4, 5, 6, 8, and 9.) Here in this problem, log(base 2) 3 = log(3)/log(2), and a quick division of these values gives about 1.585, which is between 1½ and 1⅔. No calculator, old school.

[@AmarSingh-bm3dd]

Yeah man I did the same way and I was wondering am I the only one who used the property of logarithm and calculated the exact value😂😂

[@anandsuralkar2947]

In my chemistry class we learned values of Log2=0.3 Log3=0.48... S9 log3 /log2=0.48/0.3= 48/30=1.6

[@tsoochi]

actually, log10(3) = 0.4771, and hence the answer is around 1.585 ur answer is pretty much accurate though

[@vitorangelo8502]

So it should be letter c.

[@williamjust]

I learned log(2) = 0.3010 and log(3) = 0.4771 (to base 10) from using log tables in the 1970s. And from those you can easily work out log(4), log(5), log(6), log(8) and log(9).

[@Horcrux1997]

You didn’t learn them, you memorized them. Not so practical in the long run

[@williamjust]

@@Horcrux1997 - "learn" has more than one meaning, one of which is "memorise". Actors learn their lines, for example.

[@GraemeMcRae]

Another way, which seems quicker to me, is to notice that the common denominator of all the multiple guesses is 6. So just find log_2 of 3^6 = log_2 of 81*9 = log_2 of 729, and since 729 is between 2^9 and 2^10, log_2 of 3^6 is between 9 and 10, so log_2 of 3 is between 9/6 and 10/6, which is multiple guess b.

[@1trucxhondamov589]

This is interesting! I gotta maker this! Glad I subscibed!

[@anshuagrawal7364]

At first I thought it was a complex problem, but after the explanation I realized how easy it was! Thanks for the great explanation!!!

[@vinitshandilya]

I calculated the lower bound as 2e1.5 = 2*sqrt2 = 2.8, which is less than 3. You just need this piece of information to eliminate all the remaining choices, except (b). There was no need to calculate the upper bound. :)

[@siddharthmishra1061]

I'd be really happy if this question came in iit jee.

[@baharat4315]

Same, i am not in india but these questions they ask in the west seems really easy comparing to asia.

[@choerrim7745]

YanAli YanÖzgür Oxford isn’t in the west.

[@milikumari5474]

Nope not really dude.. Coz easier the paper, higher the cutoff

[@milikumari5474]

@Muhammad Qasim you got a point man

[@deepthink5397]

munna , itta golu molu wala question nhi aayega

[@nihalbadiger8922]

We can calculate the actual result. Do this: (log3 to base 10)/(log2 to base 10) You get (0.48/0.3)=1.6

[@oleum5589]

Exactly lol this was so easy

[@mokshgupta6110]

Yesssss. I don't know what other people on comment box are on.

[@oranbrie]

how would u calculate that without a calculator?

[@maniebrahimi7952]

I swear if I was to do it, I’d do everything right and then confuse the sings and put the wrong answer

[@NOMATfull]

So my answer is "more than 2 but less then 3/2"

[@astroknight5]

The fact that you got confused in the spelling of signs too, just makes this comment better.

[@vashon100]

@@NOMATfull than vs then

[@NOMATfull]

@@vashon100 good

[@YuLum]

just crossed out the words “no calculator”

[@DiptarunChatterjee]

❤️❤️ today i felt love with log series.. i hated it when i was learning

[@Amitkumar-ss4fo]

When I saw the thumbnail I thought we have to find exact value

[@紫瞳-w6t]

log10(2) ~ 0.3010；log10(3) ~ 0.4771；logA(B)=log10(B) / log10(A)；So, log2(3) = log10(3) / log10(2) ~ 1.585 。

[@nick46285]

就跟你講不能用計算機了

[@紫瞳-w6t]

@@nick46285 在我念高中的時候3個常用的對數值是基本常識之一。 log10(2)~0.3010, log10(3)~0.4771, log10(7)~0.8451。 log10(1~9)都可以由此3個值來推導。

[@nick46285]

@@紫瞳-w6t 啊這就是用了計算機你才知道啊

[@yichen6313]

我高中的時候也背過！現在如果考試突然碰到的話的確是不用計算機⋯

[@nick46285]

@@yichen6313 你到底聽不聽的懂人話，你不用計算機能自己用手寫的把對數值算出來????靠背的就已經不是用筆算了，而是用計算機算完自己背，你背教科書一樣是教科書作者先用計算機算出來

[@e8heterotic649]

If you know some interesting music theory you can answer this quickly. Modern tuning is based on 12 equally spaced notes per octave. One octave is a doubling of frequency, a ratio of 2:1. The perfectly tuned fifth is a frequency ratio of exactly 3:2. However, in 12 tone equal temperament, the fifth is flattened ever so slightly so that 12 fifths lines back up with the octaves. Nonetheless, the 3:2 ratio is still very close in our tuning system. In 12 equal, 12 fifths should be the same as 7 octaves, meaning that we pretend the following equation is true (which it approximately is): (3/2)^12 = 2^7 which means: 3^12 = 2^19 This means that the base-2 log of 3 is approximately 19/12. There are even more accurate tuning systems (for fifths, at least), like 53 tone equal temperament which is often used in traditional Turkish music. There, an octave is 53 steps and a fifth is 31 steps. This gives: (3/2)^53 = 2^31 or 3^53 = 2^84 Thus, log_2 of 3 is about 84/53. This is much more accurate than 19/12, but even 19/12 is good enough to answer this question. You can find increasingly better tuning systems for the fifth by looking at continued fraction approximations to log_2(3), of which both 19/12 and 84/53 are parts. Notes per octave is in the denominator, the numerator is how many notes in an octave plus a fifth.

[@angelmendez-rivera351]

This is good and all, but this all takes more mathematical work than what was shown in the video, so it's not exactly the best thing to use. It's only convenient if you're infinitely more skilled with music theory than with basic arithmetic, but this is extremely unlikely since music theory, in a meaningful sense, IS just arithmetic. Someone who is good at music theory would probably not be bad at arithmetic.

[@phenomenal5743]

I took a different approach: 2^x = 3 x >1 and x < 2 { 2^1=2 and 2^2= 4} So 1

[@Beerfazz]

2 sqrt 2 is ~ 2.8

[@lecinquiemeroimage]

artificial and very bad solution ! (please have a look on my "post for all").

[@robinder_]

yeah fam same did it in like 10 seconds lmao, high iq gang

[@phenomenal5743]

@@robinder_ haha😂😂

[@anandk9220]

The concept of lower and upper bound used here is excellent and amazingly good to be applied in such a problem. Truly amazing to solve without calculator! 👍 EDIT : By the way, I already know logarithms of 2 and 3 each to the base 10, which are 0.3010 and 0.4771 respectively. So I'd simply divide these values and make further adjustments to calculate the answer. This approach is definitely not worth the Oxford level, but I think answer will be correct. 😊

[@pat1x179]

If you know the Graph of ln(x) u can calculate: ln(3)/ln(2) because logx(y)=ln(y)/ln(x) and then you have to guess what number u have to put in, ln(pi)=1 so that means that ln(3)=something like close 0.95and ln(2) maybe about 0.2 or 0.3 smaller than ln(3) so u calculate 0.95/0.7 or smth like that and you should get a number thats in the right range of the solution

[@jerrysstories711]

I did it a cruder way, perhaps more realistically the way one would do it under time pressure. From memory, 2^(1/2) = 1.41… therefore 2^(3/2) = 2.82… From some quick pencil work, (5/4)^3 = 125/64 < 2 therefore 5/4 < 2^(1/3) therefore 25/16 < 2^(2/3) therefore 50/16 < 2^(5/3) QED 2^(3/2) < 3 < 2^(5/3)

[@biogoo]

I usualy try to solve your problems from thumbnails only before clicking on the video. I was wondering how exact do I have to be. My solution was log(3) = log(2 x 1.5) = log (2 x 1.05 x sqrt(2)) = log(2) + log(sqrt(2)) + log(1 + 0.05) = 1 + 0.5 + 0.05 = 1.55 With 1.5 being about 5-6% off square root of 2, rounded down to 5% as we are in logs and the approximation of log of small number from Taylor is above the true value. Then I clicked on the video and saw I have 4 options...

[@veyselyazici]

biogoo can you please explain log(1.05)=log(1+0.05)=0.05 part?

[@biogoo]

Well, it is not very accurate, but I basically approximated a log of a base 2 with a natural logarithm and then approximated the natural logarithm with the first term of its Taylor series (natural logarithm of 1 plus a very small number is almost equal to the very small number). If done properly, the Taylor series of a natural logarith should be multiplicated by 1/ln(2) to make it a Taylor series of a logarithm of a base 2. Now the question is what the hell is ln(2), but if one would approximate e = 2 x sqrt(2), which is less than 5% off, then solving ln(2) = X is like solving 2 = (2 x sqrt(2))^X 2^(1/X) = 2 x sqrt(2), with 2 x sqrt(2) = 2^(1.5), meaning that 1/X = 1.5, which means that 1/ln(2) can be approximated with 1.5. So, a better approximation would be log(0.05) = 0.05 x 1.5 = 0,075 :)

[@jacknguyen5220]

Should've divided the last bit by ln(2) (as you point out below). A better rough approximation is to use ln(x) ≈ 2 (x-1) / (x+1) log2(3) = 1 + log2(3/2) = 1 + ln(3/2) / ln(2) ≈ 1 + (2/5) / (2/3) = 1 + 3/5 = 1.6

[@jacknguyen5220]

@David Schmitz Yeah I definitely don't have powers of 2 or 3 memorized to 2^19 and 3^12, and think neither do most people... Getting better estimates from my method is not too bad, especially if one knows ln(2) in advance or has a calculator for basic operations.

[@groszak1]

@David Schmitz 41log₂x is also accurate up to 16 0 41 65 82 95 106 115 123 130 136 142 147 152 156 160 164

[@abdullahcelik35]

This is a regular high school maths exam question at a over average Anatolian High School in Turkey LOL.

[@Mn-Fe-N]

No choices needed: I solved the problem *by music* We know that 12-equal temperament is an approximation that 7 octaves ≈ 12 pure fifths, which means 2⁷ ≈ (3/2)¹². So 2¹⁹ ≈ 3¹², log₂3 ≈ 19/12 ≈ 1.58. For a more accurate answer: 53-equal temperament is a better approximation that 31 octaves ≈ 53 pure fifths, which means 2³¹ ≈ (3/2)⁵³, so the answer is (53+31)/53 ≈ 1.5849. Precise value: log₂3 ≈1.58496, the estimation is nearly perfect.

[@groszak1]

yeah I know the 84/53 approximation I also know this: x = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 41log2(x) ≈ 0 41 65 82 95 106 115 123 130 136 142 147 152 156 160 164

[@Mn-Fe-N]

@@groszak1 yeah, 41 is also a good approximation, but not as well as 12, 53 and 665. But why do you want to remember these?

[@groszak1]

@@Mn-Fe-N because, 41log2(x) approximates not only 3 but also 5, 7, 11, 13

[@devkumar9889]

What is this temperament ??

[@DunderBallZ]

Damn Asians

[@trash9598]

i literally watch these at like 3 am having no clue what hes talking about lmaooo

[@IDMYM8]

Your profile pic says it all

[@communist4life630]

You are too young

[@markspazzmaster388]

SAME

[@factory_enslavement]

@@elvir182 then you're an american

[@technicalmaster-mind]

Exact 3:00 AM in my phone.. LoL!

[@sujalagarwal9932]

Preparing for JEE ...I think my approach would have been very simple like log 3 base 2 =log 3/log 2=0.4771/0.3010....and we can calculate it in our mind that it is somewhat greater than 1.5 and less than 1.6 .. This would have been my approach... basically in JEE we have to remember the values of both log 3 and log 2

[@j.u.4.n620]

Yes absolutely r8

[@pedroxavier4227]

The kind of proplem you know in theory, but do not have the balls to do in practice

[@davemwangi05]

a lazy choice. LOL

[@pedroxavier4227]

@@futilitariano don't we all know a girl who can this exercise? Im reaching out for the majority of youtube people to make my comment more relatable and enjoyable.

[@smitajky]

@@pedroxavier4227 Why a GIRL? Of course I can and do this sort of thing many times every day and have for 6 decades.

[@pedroxavier4227]

@@smitajky cpuse the first guy said a girl

[@daddy2513]

Stfu and go do ur homework

[@keedt]

Weird solution method: if you are into music theory, you know that the interval of 3:1 corresponds to an octave plus a perfect fifth, which in 12-tone equal temperament corresponds to 12+7=19 half steps. Hence 2^(19/12)≈3, and as 19/12 is slightly over 3/2 we find the right answer.

[@keedt]

@@easymathematik The observation I made was that if you know music theory, you also know a (quite crude) log_2 table without probably realizing it.

[@easymathematik]

@@keedt Sorry for the mistake. :)

[@slidetapgames7273]

I didn't watch the video before trying the question so I thought you had to just find an approximation, which you can do by using the change of base rule to write it as ln3/ln2 or equivalently ln(1/3)/ln(1/2) and then take the first few terms of the Maclaurin expansion of ln(1+x) to get an approximation.

[@speedseeder]

'solving' is a strong word for this guesswork

[@johnchessant3012]

Reminds me of another interview question: which is larger, log_2(3) or log_3(5)? Answer: We see that log_2(3) > 3/2, while log_3(5) < 3/2. Hence log_2(3) > log_3(5).

[@targetiitbcse1761]

Nobody: Presh: Starts taking questions from quora

[@hritanshurath4353]

When you have been watching mind your decisions so much that this seemed really easy even as a high school student

[@ryannoonan5518]

Draco Dasher obviously multiple choice isn’t hard I don’t think it’s because of watching them lol

[@AK-gb9rp]

I'm 3 and I solved this with my ears closed

[@denizkaragullu6239]

@@ryannoonan5518 mutliple choice definitely helps a lot. I just tried the choises lol

[@choerrim7745]

The multiple choice makes it so easy that even an eighth grader can solve it *cough*

[@austincheng642]

Anyone who knows the basics of logs and exponents can solve this easily

[@cardinalityofaset4992]

You can actually solve it faster if you use your imagination

[@jibran8410]

right but oxford is looking for people who use upper bounds rather than approximations and are able to adapt to a problem.

[@ramone.chacon5084]

Its basicallly like using a calculator.

[@DarkMatter1919]

Then get it wrong

[@fabriciaoliveira3236]

@ehud kotegaro you can try to solve it imagining a graph

[@betterbegood.]

@@fabriciaoliveira3236if you remember the frist decimal of sqrt(2) this problem is easy

[@karabo506]

I always choose option B when i don't know the answer, and it's been working out very well so far

[@datphung8656]

Did you cheat? A. No B. Most definitely

[@moonexpr]

I just calculated 2¹ and 2², 3 is between 2¹ and 2² and we know it's definitely not 2 so that narrows down our exponential to be somewhere (1, 2)

[@Dubanx]

I took it one step farther. If 2^1=2 and 2^2=4 that means the average slope is 2. If we drew a straight line it would intersect 3 at X=1.5 That said, it's not a straight line. It's a curved line which means the slope between X=1 and X=1.5 is BELOW the average and, therefore, 2^1.5 is BELOW 3. So clearly X>1.5 The tricky part is determining whether it's greater than 1+2/3 or not.

[@crashedg7976]

I had a dream of eating a marshmellow. When I woke up, my pillow was gone

[@crashedg7976]

No and no

[@4919abcd]

In Taiwan, students in high school need to memorize the value from log 2 to log 9. So this will be a simple problem if it is a selective problem.

[@groszak1]

41, 65, 82, 95, 106, 115, 123, 130 in units of ¹/₄₁

[@vats2303]

That's to the base of 10 This is to the base of 2

[@4919abcd]

@@vats2303 But log3/log2=log2 3

[@vats2303]

@@4919abcd Uh... I'm not sure if that's correct..You can check that somewhere?

[@4919abcd]

@@vats2303 Trust me. That's correct.

[@user-jm5xw3wg6g]

In Taiwan,log2,log3 are the common estimated number we have to memorize. So that log2‘3=log10 3/log10 2=about1.58

[@tsoochi]

same, too easy for Taiwanese student

[@user-jm5xw3wg6g]

@@tsoochi Hello

[@AlekVen]

Logarithmic function is decelerating, so the middle point between two whole numbers of that function would lie over the respective middle point of the argument. You don't need any calculations to figure out the lower bound of 3/2. The upper bound though you gotta do like in the video.

[@НиколайЧуприк-ъ4с]

Of course. Its the most obviously. I hope such comment will be top.

[@CaoNiMaBi]

before watching: between 1 1/2 and 1 2/3 because 2 to the power of 1 1/2 or 3/2 is 2 x root 2, which is 2 x 1.4something, which is a little below 3.

[@nathanisbored]

for some reason i was thinking 'no calculator' meant 'do everything in your head', so i was trying to divide ln3 by ln2 in my head numerically and i was like ok CLEARLY im doing something wrong

[@wwoods66]

Right? Everybody knows log(2) = .3010, but who learns log(3) ?!

[@brianbrain6125]

2^2 = 4 > 3, eliminate (d) 2^(1.5*2) = 2^3 = 8 8 < 3^2 = 9, eliminate (a) 2^(5/3 *3) = 2^5 = 32 3^3 = 27 < 32, eliminate (c) so the answer is (b)

[@europebasedvlogs1251]

Ditto

[@stavats]

This

[@abomohammedalmashhrawi8527]

Good ! But if there no choices جيد ولكن إن لم يكن خيارات!

[@easymathematik]

@@abomohammedalmashhrawi8527 U would do the same. U look for the nearest power where u know the exponents. And one could be a little bit cleaner. log_2(3)=x 2^× = 3 A bountiful estimation could be to take a look for the nearest powers. a) 2 = 2^1 < 2^x < 2^2 = 4 (2^x is monoton increasing, therefor the inequality) b) Since 2^x is monot. incr., log(x) is mono. inc. c) log is continious (very important) All together: there is a unique x between 1 and 2, s. t. 2^x = 3 For better estimations u have to be "smart". "Smart" in the sense to play with powers, for example. 2^3 = 8 < 9 = 3^2. This 9 is the gag. U can use also 2^3 = 8 < 25 = 5^2. Why not? But this estimation is very bountiful. A smaller upper bound is better. So u play with powers in this case.

[@alexchou1984]

Considering the options, we can simplify this to a comparison between 2^4 vs 3^3 , 2^3 vs 3^2, 2^5 vs 3^3, and 2^2 vs 3^1, and observe where the inequality switches directions. So 16 < 27, 8 < 9, 32 > 27, and 4 >3. The switch occurs between 2^3 vs 3^2, 2^5 vs 3^3, putting the answer between 3/2 and 5/3.

[@ketankunkalikar2682]

Couldn't we have applied change of base theorem and estimated?

[@andreamusso1469]

Yeah. His method is a bit "cleaner" though

[@logician1234]

I see you like to live dangerously

[@razean22]

thought of the same but before admission its unlikely you know it. at least in germamy.

[@azamabdullatiff6054]

What base u gonna use then?

[@balrogrex]

yes, that is what i did (make it base 10) to get the lower boundary of 3/2. his method is better but he does not explain it well. like why raise to 3 and 2, how did he come up with those numbers.

[@Driftwave_Beats]

2^1.5 is 2*sqrt(2) which is a little more than 2.8 2^1.66 is 2*cbrt(4). cbrt(4) is greater than 1.5 so this is greater than 3. So the answer is b

[@joaoluizmoraesgomes7778]

I thought exactly like that

[@transklutz]

The solution does not need one to remember or calculate square or cube roots.

[@micka6288]

Exactly the way I got it

[@babbintandukar9959]

Ah we simple minded know how to solve but we still fail cause in exams we have to solce by specific process

[@lordlix6483]

I used the same logic :O

[@droopyphantom6136]

It was really an easy one:- Log3 base 2 can be written as log(3/2) i.e 0.4771÷0.3010 which is equals to 1.58 approx and then with the help of option we can just select the right answer

[@catarinacarvalho7959]

Awesome. Great channel, by the way. Should be exist more like you. Greeting from Portugal.

[@grobertoac2430]

I saw the thumbnail, and thought... 2*sqrt(2) ~= 2.8 ~ log,2(2*sqrt(2)) = 1.5. So, log,2(3) ~= 1.6

[@matko8038]

Same

[@jimjab3631]

Same also, but faster

[@lordarty3315]

for a non multiple choice approach, i did it this way: 2^11 is aprox 3^7, log both sides and get 11log2 aprox 7log3 and finally: log_{2}3 is aprox 11/7. (11/7 is 1,571... and log_{2}3 is 1.5849..., so its really close for a quick approach)

[@andeheri]

I just wanted to say that you are such a huge inspiration for me, and I simply love watching your videos! Keep up the great work!

[@starkendeavours7072]

Thanks! for this question, now I am feeling that most of the things are analytical in mathematics. The perfect one!

[@twks123]

This vid is published on 19th of year 2019, with length 3:19. At the time of this comment, the number of views ends with 19. 19 is the deepest level of hell just for your soul (even deeper and worse than 18th level of hell most ppl knew). In the 19th level of hell, your soul will be tortured in the worst and the most painful way ever among all past and future souls ever existed, with this torture on your soul at 19th level of hell be forever nonstopping and lasting forever to infinity!

[@starkendeavours7072]

@@twks123 Whether a soul will go to hell and at which level of hell OR in heaven is being decided by his actions at this very present moment of his life. The more torture( if it belongs to infinity), the more powerful will be the tolerance level of every Human being and it tends to the Cosmos Eternity, invincible Level

[@cholifahfitria8947]

OMG, I just taught my students today about this! We learnt about drawing the graph of logarithm functions and when we wanted to decide what value of x that should be used in the function, I told them to find a value that make them easier to count without calculator. And I said, like Log2(3) is hard to find the result unless you calculate it with calculator. BOOM, now this video was in my reccommendation list. WHOAAA thank you! :D

[@Imna_Jamir_]

Others: *uses theories I never learned* Me:Hmm..I picked no."B" last time so I'll go with "C" this time

[@communist4life630]

Not only its a terrible way but also picking different answers makes your chances of picking right lower

[@jagadeeshjagadeesh8545]

Nice one😂

[@aditikulkarni8552]

Damn i was thinking of the exact value from the thumbnail 😂

[@mohammedshahrukh8773]

Same here

[@irrelevant_noob]

ikr, i was considering how to get enough decimal places so that my answer would be acceptable enough... :-)

[@sakshamraj1333]

Btw I got the exact value simply without calculator

[@irrelevant_noob]

@@sakshamraj1333 oh have you now?! What's the 97th decimal then?

[@KorZen10]

@@sakshamraj1333 Oh, good on you. Mind showing us how?

[@el9418]

Another solution is to graph the function y = 2 ^ x. It passes through (0,1), (1,2), and (2,4) and takes the general arcing shape of an exponential curve. It's easy to eyeball that the curve would pass through y = 3 somewhere a little bit higher than x = 1.5

[@roshan459]

Me after seeing the Thumbnail: log2 3=x 2^x=3. 2^1=2 & 2^2=4. => 1

[@manthanc7727]

Still Good, doubles your stakes!

[@theimmux3034]

So you just knew that 2^(5/3) is greater than 2? If yes, then you could've just gone your way and noticed that 2^(3/2) is the higher bound of a) and that it is equal to twice root 2 which is about 2,828. The higher bound being less than 3, b) has to be the correct answer.

[@nybotheveg]

I remembered that sqrt of 2 is 1.4..., so 2^1.5=2*1.4=2.8..., so I figured it had to be slightly above 1.5.

[@justheretocomment7318]

When in doubt go with C

[@integralboi2900]

What should my integration constant be? C, of course.

[@glitched_matrix44]

@@integralboi2900 yeah....

[@Milkaerys]

What type of Hepatitis do I have? Probably C...

[@SergeyVlasov_io]

Doc, which vitamin supplements should I take?

[@SergeyVlasov_io]

Which programming language is both old enough and still very popular?

[@luisbenites4825]

Not sure if this has been shared below. As some have said, by inspection (or following the same argument below) it's either B or C. So without using the calculator we just need to know if 2^(1+2/3) is larger than 3 or not: 2^(1+2/3) = 2x2^(2/3) = 2x4^(1/3) So we just need to determine if the cube root of 4 (the 2nd term above) is more or less than 1.5 (given that 2x1.5=3) : Since 1.5x1.5x1.5= 2.25x1.5 > 3, then the cube root of 4 must be less than 1.5, so the answer is B

[@vickysingh9736]

I remembered the log values(from iit jee prep 6 years ago) of 2 and 3 so I just divided it and got the answer. Lmaooo

[@vibhudixit1336]

Same here bro I just passed the jee mains exam and in coaching they always told us to remember these values

[@sakshamraj1333]

Me too

[@ramazanozdemir1200]

Lol am from turkey thats so ez.

[@devd_rx]

lol same, but i am 10th grader

[@ompatel6819]

Hey Presh!! You can also use 'Base Changing Theorem' of log to solve this question easily.....!!

[@jonasdaverio9369]

What base would you chose?

[@vaidehibaranwal9245]

@@jonasdaverio9369 10

[@starpawsy]

If you write down powers of 2 and powers of 3 and find "similar" values, you can get a lot closer. So if you have 2^a ~= 3^b and take log base 2 on both sides, you get "a" ~= b log3. Thus, log3 base2 =~ a/b. For example 2^11 = 2048 < 3^7 = 2187. So log 3 > 11/7 ~= 1.57 3^5 = 243 < 2^8 = 256. So log 3 < 8/5 = 1.6

[@WelpNathan]

I got lost when he said "log."

[@jachymdolezal5103]

you don't understand what a "log" means? You won't be a good lumberjack then ...

[@fernandocontreras4981]

@@jachymdolezal5103 **Badum tss** not gonna lie, it gave me a chuckle

[@adrianpramadipta8631]

Jáchym Doležal hahahaha

[@pointlesslylukesplainingpo1200]

logarithms are just another way of writing exponents/powers e.g. log(x)8=3 would just be x^3=8, 2^3=8, so x=2

[@nextbigthing2917]

The DeadCreator you lost me at the first word of you’re comment.

[@audriusbabraitis4826]

Why did he took 3 and 2, not the other numbers?

[@Leon-ge4tu]

Because it was enough to solve the problem

[@theloganator13]

Behold, the art of knowing the answer before you try to solve the problem.

[@justbram5198]

Smaller numbers are easier to calculate.

[@yurenchu]

If he had chosen 8 instead of 3 and 2, then he could have solved it with just one calculation. (2^x)⁸ = 3⁸ = 6561 ==> 12 < 8x < 13 ==> 12/8 < x < 13/8 ==> 1.5 < x < 1.625

[@binaryagenda]

Because the bounds in the question had denominators of 2 and 3. If the multiple choice options had other denominators those might have been needed too.

[@VitoChouMalk]

In taiwan, high school student are required to memorize the approximate values of 3 useful constant: log2=0.3010, log3=0.4771, and log7=0.8451. So this question become simple as we know log_2 3=log3/log21.5

[@konuralpyldzkan1495]

We can also use a graph. Even with a handmade graph, answer can be easily seen

[@Xelianow]

We can rule out Option D because 2^2 is already more than 3, therefor it can't be between 2 and 3. The graph is concave, which means you can easily see that the answer has to be greater than 1 1/2. But a graph does not realy help whether the answer is less or greater than 1 2/3, thats where you run into a problem...

[@sean_haz]

@@Xelianow a graph could give you am approximation of the answer which would fall between only 1 of the ranges

[@transklutz]

@@sean_haz no need to approximate when you can be precise like he has done.

[@Xelianow]

@@sean_haz Yea, but how do you get that graph without a calculator? You do have some nice points you know of, like (1;0), (2;1) and (4;2), but apart from that you have nothing to work with except the concave curvature and the general shape... But without a calculator you do not have any idea whether 3 is above or below 5/3... at least not until you calculate it, in which case we are in the same situation as befor^^

[@sean_haz]

@@transklutz i agree his solution is better, but i think this solution would also work given that it's multiple choice

[@userBBB]

so they expect you to solve course level question before you take the course?

[@henryginn7490]

BBB they design the papers so you can do them using only stuff taught in the first year of A level maths

[@williamjust]

They expect you to show you can use what you've been taught at A-level intelligently, rather than doing it by rote, (which I suspect is how maths is usually taught these days, judging by the number of complaints when an exam question requires the application of more than just one basic technique).

[@jashan_iitroorkee]

We can also find the exact value if we know the values of log (base 10) for numbers 1 to 10. Log3(base 2) can be written as log3(base 10)/log2(base 10).

[@Saimiri1334]

I've always wondered in problems like this, why you are able to take '32' and '8' in the inequations. Is this a smart guess since 2^3 and 2^5 are higher / lower than (2x)^3, if x has to be >= 1? I don't see a concrete rule there :(

[@mario_gabriel]

Saimiri Ray because he just needs to find the limits, you know 2^x has to be 3, so it has to be sqrt(9), and if you take 3/2 you know is the same as sqrt(2^3) that is sqrt(8), so x has to be bigger than 3/2 and then he took 5/3 that is the same as 3root(2^5) and 3root(32), and again 2^x has to be 3 so if you put it in 3root it has to be 3root(27), wich is smaller than 3root(32). Basically he took 32 and 8 from the given limits (3/2 and 5/3) and tested if the result was bigger or smaller than 3. In my opinion this kind of questions are made to measure your problem solver skills, everyone can follow concrete rules but not everyone can come out with his own solution to any problem.

[@Saimiri1334]

@@mario_gabriel thank you. That's what I mean, it's about testing. :)

[@TallinuTV]

Wow, I don't have a clue how you knew to choose those numbers to add as exponents and such, other than "already knowing the answer" lol... Guess this explanation is geared toward people who are already studying the stuff that teaches you how to figure that out!

[@SadisticNiles]

He very likely tested 2 to the power of 1,33; 1,5; 1,66; and then after he figured it out he just presented the answer in a way which is very different from how you actually approach it (likely because it's more neat this way).

[@angelmendez-rivera351]

You don't need to already know the answer. All you need to know is there exists a pair of rational numbers a and b such that a < x < b, hence 2^a < 2^x < 2^b. In the case of 2^x = 3, it is notable that 2 < 3 < 4, and 2 = 2^1 and 4 = 2^2, hence 2^1 < 2^x < 2^2, hence 1 < x < 2. However, this does not give you the answer. Then how about squaring the equation? 2^(2x) = 3^2 = 9. In between what powers of 2 is 9? Between 8 and 16, and 8 = 2^3 and 16 = 2^4. Hence 2^3 < 2^(2x) < 2^4, hence 3 < 2x < 4, hence 3/2 = 1 + 1/2 < x < 2. This still not quite the answer you want, though, so let us try the next power: cubing. And the reason you know you want to square and cube these numbers is because in the multiplie choice question, all the answers either have 2 or 3 as the denominator, which corresponds to sqaures and cubes. Hence you consider 2^(3x) = 3^3 = 27. Between what two powers of 2 is 27 in? Between 16 and 32. 16 = 2^4 and 32 = 2^5. Hence 2^4 < 2^(3x) < 2^5, hence 4 < 3x < 5, hence 4/3 = 1 + 1/3 < x < 5/3 = 1 + 2/3. This is still not quite the answer, but at least now you have two inequalities. The first is 1 + 1/2 < x < 2, and the second is 1 + 1/3 < x < 1 + 2/3. Next, you notice that 1 + 1/3 < 1 + 1/2 and 1 + 2/3 < 2. Hence 1 + 1/2 < x < 1 + 2/3. This agrees with exactly one of the multiple choice answer. Hence you can stop here.

[@austincheng642]

It's actually a pretty simple concept. You plug in the values and see whether it is above or below 3. While it at first may appear hard to do numbers like 5/3, that's actually just equal to the cubic foot of two to the fifth.

[@kiwi5792]

log base 2 of 3 = "two raised to what power equals 3" 2^(5/3) = cube root of 32, which must be greater than 3 because 32 > 3^3 2^(3/2) = square root of 8, which must be less than 3 because 8 < 3^2 B, its between 3/2 and 5/3

[@chang4852]

because log base 2 4 is 2 and log base 2 2 is 1, log base 2 3 is like basically 1.5 so i just guessed B no?

[@zcherradi2]

Nope it isn't working this way

[@stealthemoon8899]

I solved it by thinking 2^x must be 3. That means 2^x has to be 2*1.5. I found 2^(3/2), or 2^(1 1/2), is 2*sqrt(2), and sqrt(2) is just under 1.5. 2^(1 2/3) is over 3, so b had to be the range.

[@smitajky]

I just used linear interpolation and numeric techniques. It took only 2 seconds to get the approximation 2^1.6 ~= 3 ( 2^1.5 must be closer to 2^1=2 than 2^2=4) It didn't need anything like the complexity of the proposed solution. The thing is that numeric methods can work for any type of function or expression.

[@moskthinks9801]

To be honest, I remember this value from 3b1b fractal video :), and had seen the solution on bprp.

[@azureabyss538]

You're amazing on Quora. Nice to see you here.

[@moskthinks9801]

@@azureabyss538 Thanks!

[@kk-xp2fj]

Can i ask why you chose power 3 to find the upper bound and power 2 for lower bound. Can someone explain this to me?

[@wangxiuying6018]

Yeah.. I dont get it

[@ayyubshaffy3612]

this question requires trial and error.lets say u chose 1 as upperbound then u get the inequality x

[@CarrotCakeMake]

He actually did trial and error with all the answer bounds, found the correct answer, then reverse engineered this solution based on the known answer. It is trivial to check whether $2^(1+1/2) > 3$ and $2^(1+2/3) > 3$ just using simple arithmetic, he just didn't want to admit that was what he was doing.

[@yurenchu]

@Kerwin Ng , I'm not sure if Presh was aware of this, but the way to solve this question is to use the denominators of the bounds in the given answer options as exponents. It's even quicker if we'd use the _least common multiple_ (LCM) of the denominators, because we could then find the upper bound and lower bound simultaneously: Let x = log₂(3) , then 2^x = 3 The denominators of the bounds in the given answer options are {1, 2, 3} , and LCM(1, 2, 3) = 6, so use 6 as the exponent to which we raise both sides: (2^x)⁶ = 3⁶ = 729 , which lies between 2⁹ = 512 and 2¹⁰ = 1024 , therefore 2⁹ < (2^x)⁶ < 2¹⁰ 9 < 6x < 10 9/6 < x < 10/6 3/2 < x < 5/3 ==> the correct answer is option (b). As another example: Suppose the five given answer options were: p) between 7/5 and 10/7 q) between 10/7 and 11/7 r) between 11/7 and 8/5 s) between 8/5 and 12/7 t) between 12/7 and 9/5 The denominators of the bounds in the given answer options are 5 and 7. We could raise to the power of 35 ( = LCM(5, 7) ), but calculating 3³⁵ would be a bit too difficult. So instead, let's find upper and lower bound separately: (I) Raise both sides to the power of 5: (2^x)⁵ = 3⁵ = 243 , which lies between 2⁷ = 128 and 2⁸ = 256 , therefore 2⁷ < (2^x)⁵ < 2⁸ 7 < 5x < 8 7/5 < x < 8/5 1.4 < x < 1.6 (II) Raise both sides to the power of 7: (2^x)⁷ = 3⁷ = 2187 , which lies between 2¹¹ = 2048 and 2¹² = 4096 , therefore 2¹¹ < (2^x)⁷ < 2¹² 11 < 7x < 12 11/7 < x < 12/7 1.5714... < x < 1.7142... Combining the results of (I) and (II) gives us 11/7 < x < 8/5 , and hence option (r) was the correct answer. I hope that helps.

[@garthreid355]

I did this question another way before looking at your solution and got the same answer. I converted the log to ln and used the y = ln(x) graph to approximate the answer and I got 1.67. Hence B is the answer

[@PriyanshuSingh-uo6dr]

I did it in my mind because I knew the value of log2 and log3 for base 10 so I just converted it to log3/log 2

[@jobretten5696]

I converted it to log3/log2 but didnt know how to solve it from there. Apparently this was some analysis stuff I didn't know about, great to learn this

[@kriddusitsin7462]

Job Retten You cannot solve from log 3/log 2 but since most people who know logarithm also know the (approx.) values of log 3 (0.4771) and log 2 (0.3010), they can replace log 3/log 2 with 0.4771/0.3010 with is roughly 1.6 which falls in between 1 1/2 and 1 2/3 so the answer is B).

[@jobretten5696]

@@kriddusitsin7462 I know, I didn't know what the approx answers of log3 and log2 were so I couldn't continue.

[@育銘-z3e]

log3=0.4771 log2= 0.3010 we have to memorize them until log9 in taiwan

[@vedantthapar3666]

**when you get the answer but it’s so simple that you thought it was wrong** BTW I didn’t write down the options so I had no idea that it was actually right

[@Weretyu7777]

Thanks for getting us a more exact answer. It's also easy to get an answer of which integers it is between by realizing the same thing: that log_2(3) is the same as 2^x=3. So we're looking for the exponent that when two is raised to it equals three. It can't be one or less, since 2^1 is 2, and it can't be 2 or more, because 2^2 is 4. So the answer has to be between 1 and 2. Which does actually help eliminate answer D, because it is between two and three, which is not possible.

[@Billyman8]

I’m in grade 8 and I see this in my recommended 😂 But thanks anyway

[@longduytran3092]

Like me

[@future62]

I just figured 2*SQRT(2) = 2 ^ 1.5 = ~2.8, so x would have to be bigger than 1.5 but not by much. I was ready to guess 1.6

[@sshlyk]

future62 neat

[@mario_gabriel]

future62 i was ready to guess the same, but I got that in a different way, resolving 2^(3/2) you get sqrt(8) but we need sqrt(9), and the next one is 2^(2) wich is 4 that is also sqrt(16), so the sqrt(9) is much closer to sqrt(8) than it is to sqrt(16), it means it has to be around 1.6 and it even makes more sense to me that it is less than 1.6 because if you try to imagine the same scale of 1.5 - 2 between sqrt(8) and sqrt(16) then the sqrt(9) is closer to sqrt(8) than 1.6 is to 1.5, maybe 1.58 or something like that.

[@seanclough7810]

Looking at a book in real analysis after getting A's in lower div maths is the reason I'm not a mathematician. If only I had 100 of these to do after seeing you explain it. Removing the exponential (and its inverse) simplifies it to simple algebra. huh ... did not see the execution until you showed me. Thank you.

[@jeofthevirtuoussand]

I used the remaining of my soul to try to understand, there is a lot of math that I ve never been told at school , to bad it is kind of fun

[@peterk.6093]

Exactly the same for me. I was never learned logs at school, nor anything more complicated. I am just discovering it now, years later and it is quite fun. Actually looking forward to go through this with my kids.

[@jorgeeduardopereztasso6134]

These are the type of exercises that always makes me think how beautiful maths are.

[@taeromann5929]

Beautiful and cruel as well

[@MarioSpiteri]

I used a graph for y=2^x for x = -1, 0, 1, 2, 3 and a quick free-hand curve shows that the answer was "b" (1.5 < x < 1.67)

[@jirivacek8367]

2^8=256 ; 3^5=243 x~8/5=1.6

[@dalbizu]

That's exactly the way I solved it :)