How do I find x? Exponential equation with two different bases. Reddit precalculus r/Homworkhelp

[@bprpmathbasics]

This problem deceived me so hard, am I stupid? Factoring & quadratic formula, Reddit r/Homeworkhelp kzbin.info/www/bejne/oHSsl6KBlq11jpY

[@NateDaGreaty]

I have a question, is 0.99 repeating = 1? Im not sure since 1/3=0.333 repeating, that means 3/3 = 0.99 repeating which is also 1.

[@kyloren3587]

​@@NateDaGreatyYou should double check just to make sure, but I think ya, people do count it as 1 instead

[@NateDaGreaty]

@@kyloren3587 but 0.99 repeating isnt = to 1, im not asking what people say, im curious if it is right, which idk as my previous reason was stated, only 1 is = it 1

[@kyloren3587]

@@NateDaGreaty yes but 0.9 repeating is so close to one as a number that it might as well be it. Unless there is some kind of real world application that requires us to be infinitely accurate, which as of now isn't even known to exist as a thing

[@NateDaGreaty]

@@kyloren3587 Its so close but it isnt... Im not asking in real world im asking in math, 0.99 repeating is not equal to one, however 1/3 is 0.33, so forth to get 0.9999 (3/3) which also = 1, that why i confused, is 3/3 0.99 repeating or 1

[@daenerystargaryen]

I don't know why I'm watching this at 5AM since I'm a physicist doing PhD in neurophysics and computational neuroscience, but I thoroughly enjoyed this. 10/10. Younger generations are so lucky that they have someone like you explaining maths. Hopefully they'll know how to appreciate it and not waste their brains away on TikTok...

[@scienceislove2014]

Woah..sounds interesting.. can you elaborate like what things you study and tools you use?

[@johnmarcusengreso8273]

Ill take it as a compliment mr neurophysics man Im a 10th grader that likes math

[@J0EB0B555]

I'm a physics major as well but haven't taken a math course in a while. He's really helpful for keeping all the concepts fresh in my brain.

[@IskzenMisishuw]

Yes brother true

[@antonioruelas8902]

Physics undergrad here, this man (and Organic Chemistry tutor) saved me during calc 2

[@Stickman_Productions]

3:23 i got jumpscared by that 😭😭

[@masterkj0912]

Same I saw your comment just before it happened as well😅

[@reesetit]

I'm just impressed how you write with two different colors in one hand.

[@dddaaa6965]

Imagine now how sex

[@loser1233]

it's like using chopsticks

[@yafmaverick]

You must be special

[@viCuber]

Lmao exactly the same I commented about two weeks ago

[@kajalde3071]

I'm impressed that you do that with two hands

[@akifhossain5122]

I have a Zoology exam tomorrow. It's 3am. 10/10

[@ominious7082]

Yo how did it go? 😂

[@utkarshjain861]

So, how did it went?

[@tobedecided8886]

It's been 4 months. So, how did it go ?

[@0kiwwi]

​@@tobedecided8886bro never been seen again after the exam 💀

[@AbhigyaPal.9933]

​@@0kiwwihe died after that

[@keroro407]

"How to find X?" Bro, it's time to move on. Your X doesn't care about you anymore.

[@bprpmathbasics]

y?

[@TheOnlineTurtle]

@@bprpmathbasics😂 thats good

[@r0N1n_SD]

Lol. Bro got owned by Math😂

[@stalincomrade1867]

Don't you z how pointless it is?

[@Taokyle]

lol

[@MC-qm5jn]

i took precalc 4 years ago and was arbitrarily recommended this video yet I still feel compelled to do the homework this man has given

[@xxxBradTxxx]

I took it 15 years ago and I still watch these videos because it feels like a waste of effort to learn all of that an forget it. 🤷‍♂️

[@Dr.Insomniac.007]

So I'm a 3rd year medical student watching this video and I dearly enjoyed it. Its like going down the memory lane. Really smooth teaching. Kudos to you..❤

[@kaideng2571]

Hope the clerkship is treating you well.

[@Dr.Insomniac.007]

@@kaideng2571 yup, thanks. Have a good day.🫂

[@jordananderson2728]

To expand it, you use change of base to get log(96)/log(2/3). When dividing a logarithm, of course, you subtract the log of the denominator from the log of the numerator, which gives log(96)/(log2-log3). We can take the prime factors of 96: 3 and 2⁵, to get log(3•2⁵)/(log2-log3). With multiplication of logarithms, you add the logs of the multiplicands, so (log3 + log(2⁵))/(log2-log3). Finally, with exponentiation, you multiply the logarithm of the base by the exponent, which gives (log3 + 5log2)/(log2-log3).

[@jordananderson2728]

I should have used ln rather than log, but I'm so used to using log for change of base that I just did that by default. It works the same either way (:

[@GreggRomaine]

Nicely done, thanks for doing my homework!

[@paratoxicalcapybara3939]

you can just use laws off exponents bcuz that seems easier. then you log it at the end for answer turn 2^(x-5) into 2^x*2^-5 and turn 3^(x+1) into 3^x*3. an example of this is (3^2)*(3^2)=3^4 expand into 1/32(2^x)=3(3^x) do some division to isolate x as much as possible. 3/(1/32) = 96 or (1/32) = 1/96. End up with 2^x=96(3^x) or 1/96(2^x)=3^x x root everything. 2=xroot(96)*3 or xroot(1/96)*2=3 more division to isolate x. 2/3=xroot(96) or 3/2=xroot(1/96) put everything to the x power. (2/3)^x=96 or (3/2)^x=1/96 now log bcuz inverse of exponential to finnaly actually isolate x. log(base(2/3)) of 96 = x or log(base(3/2)) of 1/96 = x x= ~-11.25

[@Kasakuja]

well, how do you get from log(96) ---- log(2/3) to ln(96) --- ln(2/3) ?

[@TheWatch1]

Log(96)/Log(e) is ln 96. Divide by log e in Nr and Dr

[@ThisCreature]

Since when did i watch math for entertainment tf

[@Blrta]

Lol fr 😂

[@collinjoseph615]

Same lol

[@wrya141]

Lol

[@realGBx64]

For me, since 3brown1blue.

[@muhammetaydogmus4404]

Math is fun. Schools aren't

[@WilliamCacilhas]

7:18 this entirely depends on the calculator. I have one that has an ln key, a log key, & a key for log of any base. It’s a Casio fx-991 ES PLUS C for anyone wondering. It’s not programmable so it most likely will be allowed during test.

[@Gamert80]

I'm in 10th grade, so whenever he says "let's use this rule" I'm just like "uh huh" Edit: it's crazy how different some curriculums are in other countries.

[@Musterkartoffel]

We learned the logarithm in 10th grade😅 (Germany)

[@AutoFun_]

Why are u here?

[@Gamert80]

I enjoy watching advanced math, even if I don't understand it fully.

[@Gamert80]

@@Musterkartoffel I'm only half way through the year so I may learn it soon.

[@aaryan8104]

SAME FROM INDIA BTW

[@brown_bread_]

I gave up on maths nearly 7 years ago in school. In my post graduation i watch this and feel my antipathy towards the subject reduce a little. Thanks

[@sachinjain5367]

I Just Saw the Thumbnail And Thought " Ehhhh That looks Ez Lets Just Do It " Only to waste 30 mins And Find Out It Have Logarithm Which I Havent Studied😂

[@AlkaJha-jn6jk]

In which grade do you study now?

[@sachinjain5367]

@@AlkaJha-jn6jk 11th

[@soumyadeepdas8753]

I study in 7th grade and I know logarithm

[@AlkaJha-jn6jk]

@@soumyadeepdas8753 means you are in icse no?

[@yakuni4420]

KZbin’s algorithm is getting scarier, I’m a highschooler and was literally on a problem just like this just two days ago thank you!

[@oreivankovic7318]

bro fumbels my brain and proceedes to say:"but, here is a prettier way to do it"

[@CST1992]

C'mon dude, if you know the rules of log this is a pretty simple problem.

[@frostcrackle2374]

So maybe they don't know logarithm rules yet. C'mon dude if you can calculate a Hohmann Transfer, this is a pretty simple problem. ​@@CST1992

[@celoreads]

@@CST1992 if we know the rules of log we wouldnt be here for an explanation now would we? lmfao

[@CST1992]

@@celoreads you don't know what log is but you are on a calculus video? Go back to high school... "lmfao"

[@abcdqwerty3562]

@@CST1992Do you not realise that the title of the video literally says precalculus?

[@bprpmathbasics]

Try this next: 2^x=5^(x+2) Answer here: kzbin.info/www/bejne/jX2Qn6OJet6JitE

[@imabiggoofy]

bro really out here assigning hw 💀 (I'm in 8th grade, i dont know shit)

[@FloraLemonYT]

I actually learned this last unit. 2^x=5^(x+2) xln2=xln5+2ln5 xln2-xln5=2ln5 factor out x x(ln2-ln5)=2ln5 x=2ln5/(ln2-ln5) I’m not sure if there’s a better way to simplify it

[@IamFlaem1]

x=log2/5(25)

[@Neet_mbbs.0907]

​@@FloraLemonYTThat's correct! 👍

[@vandernight1220]

I mean: (X-5)log2 = (x+1)log3 … -> x = (5log2 - log3)/(log2 - log3) is just way less complicated than the methods shown, at least this is the standard method in uk

[@shrekyboi4776]

I solved it slightly different. I recognized that 3^(x+1) can be rewritten as [(1.5)(2)]^(x+1), which can be expanded as 1.5^(x+1) 2^(x+1). This is very helpful as it gives us an exponential of base 2 on both sides of the equation, which allows us to cancel out the x on the left side through exponent division rule. The full solution is below: 2^(x-5) = 3^(x+1) 2^(x-5) = [(1.5)(2)]^(x+1) 2^(x-5) = 1.5^(x+1) 2^(x+1) 2^(x-5)/2^(x+1) = 1.5^(x+1) 2^(-6) = 1.5^(x+1) Now we only have a single x variable to deal with, so we could simply apply log to both sides and isolate for x log[2^(-6)] = log[1.5^(x+1)] (log[2^(-6)]/log[1.5]) - 1 = x -11.257 = x

[@user-lb3ex6yh9u]

Well done

[@praneel1059]

when i saw the thumbnail i guessed that since 2^(x-5) = 3^(x+5) we can do something like 2^(x-5) . 1/ 3^(x+5) then 2^(x-5) x (3^(x+5)) ^(-1) and go on i guess . Btw im in ninth grade so i have no clue about what ln is

[@hridayjr6580]

nice is the antilog required or this is it.

[@scorpio9711]

ln is called natural log, where the base is 'e' which is called eular constant. BTW which country do you belong to

[@kashi2928]

I got that too, thanks for making me feel like I wasn't alone 😂

[@sidhaarthnair8368]

I wasn't taught log at school at all. I had to look it up online. Even though we hadn't had proper knowledge about log we still have to use in calculus

[@Roro-ej7ke]

Lol same💀

[@vogel-mc5ed]

It's really simple to solve the question. Just follow these steps: 1) change it to logarithm form 2) oh shit whats next? 3) cry 4) think about the purpose of maths later on in life 5) copy that one smart guy answers And voila, you solved the question

[@soumyadeepdas8753]

😂

[@Sayan_0-0_]

It’s 4:20 am right now and I have no idea why I’m watching this at this time. I told mom to call me at 8 and wake me up. I guess now I have a solid reason to tell her why i was awake.

[@lukaskamin755]

you can do backwords in 6:00 ONLY IF a and b are both positive (theoretically a can be 0 , but it's a disputable question)

[@amanda-we9fv]

что?

[@lukaskamin755]

@@amanda-we9fv now it's correct. I mean you can't do backwards if a and b are both negative ,roots of a and b won't be defined then, while root of ab will be defined

[@arcturusgd]

In this case it is. If it is positive that means a,b ∈ N

[@CuriousCyclist]

Thank you for taking the time to make this video. Much appreciated. ❤

[@bprpmathbasics]

Glad to help! 😃

[@lucsas9277]

the second option is always what comes to my mind first, i find it way easier and more intuitive, but ive forced myself doing the natural base method too cuz you have to know them both imo

[@riccardodellorto4267]

I do the complete opposite: whenever I see an x as an exponent, I use ln, because the calculator can eventually solve any monstrosity I type in as long as there are numbers 😂 Bringing down the x is my number one priority 🫡

[@Shiva-xp4wv]

Bro i am a Engineering major why did i click on this video

[@actionj761]

Same thing im like do i really have nothing better to do than to glance at my freshman year history 😂😂

[@yessubharv1400]

Might be 2 reason.. 1. To confirm ur solution thought process 2. Ur too dumb to be an Engineer.

[@npx_riff_lift-g]

CS major procrastinating on studying for my algorithms midterm✋🏼😔

[@imran_sholeh]

I have a question sir. Why we need to use ln instead of log, or we can use which?

[@sonvisharma5264]

Hi...we can use log in 1st method instead of ln ...i used log and the ans is same, u just need to know values of log2 and log3

[@imran_sholeh]

@@sonvisharma5264 i see.... Thank you

[@sahhanaaa]

guys we can solve it in another way too. what i did was this: i took log on both lhs and rhs. so the exponent comes down and the equation becomes like such (x-5)log 2=(x+1) log 3 now we know log 2= 0.3010 and log 3=0.477 so we just use those values in the equation (x-5)*0.3010=(x+1)*0.477 0.3010x-1.505=0.477x+0.477 this becomes -0.176x=1.982 x=1.982/-0.176 x=-11.26

[@CursiveThoughts]

It was not log tho. It's was ln.

[@sahhanaaa]

@@CursiveThoughts works w log too

[@Gaysandthechaos]

​@@CursiveThoughtsyeah you'd have to multiply it with 2.303 to convert ln to log That'd be easier ig

[@CursiveThoughts]

@@Gaysandthechaos yes

[@nothingbutpain863]

​@@CursiveThoughts, in this scenario, either 'ln' or 'log' is acceptable. This is because the bases of logarithms would get cancelled in the process as long as the bases are the same.

[@garrettbates2639]

Equivalent answer with slightly less distribution: 2^(x - 5) = 3^(x + 1) (x - 5) ln 2 = (x + 1) ln 3 x - 5 = (x + 1) log2(3) x - 5 = x log2(3) + log2(3) x - x log2(3) = 5 + log2(3) x = (5 + log2(3)) / (1 - log2(3))

[@TheEulerID]

The same way I did it, but nobody ever seems to approve of logs in bases other than e or 10 for the "official" answers, even though the end result can look simpler. Maybe it dates back to the days of log books which almost always limited themselves to natural and base 10 logs.

[@alexlu8178]

second method is very good that an algebra 2 students could solve it

[@mrtoast244]

I’m thinking of going for my masters in robotics in like two years, this might come in handy

[@livia4838]

I saw this passed by my fyp and it struck me 2 ideas on hoa to solve it, i knew the first few steps for both methods but got stuck! Thanks

[@ayushkushwaha550]

0:45 the quicker method after this step would be to divide x-5 by x+1, which would be equal to ln3/ln2. Now use componendo dividendo on both sides :)

[@joshuakohlmeyer7123]

The way he flips between pens is 🔥

[@bprpmathbasics]

Here's a video with 10 examples of solving exponential equations, from basic to hard!: kzbin.info/www/bejne/gWmmgpWJeZ6qn9U

[@thepiglin100]

Thanksss

[@im_b_ran]

As a communications student in university who hasn't learned maths in over 4 years, this video has enlightened me

[@zuhakhalid4900]

That was a really good explanation! Thank you for explaining so clearly! 👏

[@heyyythereitsjulie]

In the first example, could you have used log instead of ln? When to use log vs ln?

[@nothingbutpain863]

In this scenario, there is not difference. The only situation requiring 'ln' is when the base of an index is 'e'.

[@nothingbutpain863]

Also, ln(x) is equivalent to log(e, x).

[@DA-gs4gu]

I would consider simplify it with log to the base 10 which yields the same answer as the answer you obtained. We could write it as, X-5log(2)=X+1log(3) Which on further simplification can provide, x= -6.58/0.58= -11.3 And the answer you obtained at the end, log (base)2/3 (96)= -11.26 (approx) I feel its less hectic

[@E_velynn]

i somehow went through algebra I and II, precalc, calc I and II, yet never saw any of this and now i feel like i was robbed. this looks so interesting and i am now lamenting never having had a math teacher that makes math interesting. thanks, random math guy on the internet!

[@Roro-ej7ke]

Same story🤷‍♀️

[@ridoing9969]

The professor when ever I start copying the notes. 3:34

[@stolenmonkey7477]

That made me laugh a bit lol

[@My-yy8gz]

😄

[@advaith2977]

Just wanted to ask whether we can take ln3/ln2 = ln 1 = 0 and then solve for x with 2 cases giving x=5 and x=-1 (or ln2/ln3=-ln1=-0=0)

[@jameshy69]

australian here, i used my calculator. ive only seen the thumbnail and came straight here. the answer i got was (-ln(96))/ln(3/2) or approximately-11.257 edit: finished the video now and checked those two values of x. both were equal to my above answer. very nice 👍

[@salmanahmed5481]

I just do log on both side and got the same answer x = -11.25

[@matix2gorski984]

(⅔)^x=96 xln(⅔)=ln96 xln(2*⅓)=ln(3*32) x(ln2-ln3)=ln3+5ln2 x=(ln3+5ln2)/(ln2-ln3)

[@Dracogame]

I used to go through this stuff so easily but nowdays I forgot completely how to do it. Makes me want to study this stuff again on spare time.

[@notrishy]

Very first approach of solving exponential equations is using logarithms.

[@christopher8504]

Another way to do it like the first method that isn't exactly any faster but came to me is: once we have (x - 5) ln(2) = (x + 1) ln(3), (eq. 1) we can build h(x)=(x - 5)/(x + 1) = ln(3)/ln(2), (eq. 2) which will have the same solution despite the domain changing a bit, since the solution isn't near -1. and separate that into two equations: f(x₁)=(x₁ - 5) = ln(3) g(x₂)=(x₂ + 1) = ln(2) so solution x (to h) will be formed by solutions x₁/x₂ to f,g respectively. Which, is a linear system. now we can produce a matrix: [[1, -5, ln(3)], [1, 1, ln(2)]] which we partially row reduce to [[6, 0, 5ln(2) + ln(3)], [0, 6, ln(2) - ln(3)]] recombining, since solution to h is solution to f over solution to g, the 6's cancel and we have: x = (5ln(2) + ln(3))/(ln(2) - ln(3)) which is also the solution for eq.1 I know written out this seems long, but it went a lot faster in my head. A lot of the steps here would be incorrect if I didn't explain them carefully. Also it would probably take longer to row reduce than just doing the algebra but ¯\_(ツ)_/¯ let me know if I did anything illegal math manipulations

[@argonwheatbelly637]

I like it. 😊 It reminds me of when I use Synthetic Division to blitz through some polynomial division by hand.

[@nicholasscott3287]

Immediate reaction is "x is not positive integer, because 2 and 3 are prime, so the prime factorisation of 2^i will never equal that of 3^j, where i and j are any positive integer".

[@tobybartels8426]

This also works for negative integers, even for non-zero rational numbers. So the only possible rational solution would be if both exponents are zero (at the same time, which is not possible in this case).

[@deltalima6703]

You never know if x is a quaternion or is mod |p| or whatever in these dumb questions.

[@General12th]

​@@deltalima6703This is an algebra channel, not a calculus or analysis channel, so don't overthink it.

[@jackposiedonforever7774]

Correct

[@CMT_Crabbles]

After taking Calculus, it’s more scary seeing actual numbers! I swear all middle school maths just falls out of your head

[@omverma_1791]

Bro I'm in 10th grade and I reached (2/3)^x = 96 and was like, "Now what?". Then I realised "Oh, this is out of bounds" 💀💀💀💀

[@imagod4796]

this is 9th grade in Germany

[@omverma_1791]

@@imagod4796 I thought Asia had the toughest math.....

[@exip9288]

@@imagod4796 This is 12th grade in Turkiye (I know it sucks dumb education system) , but I learned it way before because of calc bc.

[@bruv4266]

@@exip9288 We all have shitty educations, here in Romania we learn calculus in 11th grade to 12th grade, they should have system of education like USA, this is where the people can learn it well, we have short time in school but too much to learn, cause it's not just math, its also other lesson that it supposed to be in college like physics, chemistry, etc.

[@NotKartikeySingh]

bro but if u have studied from better school in 9th they would have taught u (in india)

[@thedogwbigheaded]

I graduated from highschool this year. And I'm really glad rn because i understand what exactly he said. This is crazy for me dude

[@siddheshvispute17]

Actually the equation becomes easy, when you use log in exponential problems. Thanks ❤🇮🇳

[@slulzspot7583]

मुझे भी equation देख के वही लगा।

[@prelude8975]

I'm an international relations major and somehow watched this whole video and nodded everytime he looked at me as if im getting everything he says

[@MCFC-OK-]

oh my fricking god how many whiteboard pen boxes do you have😂

[@haku7335]

Watching a guy do math without any mistakes is so entertaining bro.

[@markosverdhi]

My friends in my old algebra class had a funny way of remembering the ln(x^2)=2lnx theorem. We called it the yeet theorem because you take the exponent and yeet that shit to the front

[@masnun_07]

I'm studying medicine. Idk why im here. 😂 but i enjoyed your content.

[@bprpmathbasics]

Thanks!

[@atharvasingh9579]

how about taking log on both sides the use the values of log2 and log3 that is 0.3 and 0.47

[@ookjannesplanting1296]

Challenge here is to find an answer algebraically/exactly, so without a calculator

[@Matthew-px4vu]

I solved it in a similar way somewhat. Started with taking the natural log but instead grouped terms like: (x-5)/(x+1)=ln3/ln2 (x+1-6)/(x+1) = ln3/ln2 1-6/(x+1)=ln3/ln2 (x+1)=-6/(ln3/ln2-1) x=-6/(ln3/ln2-1)-1 x~=-11.257

[@nothingbutpain863]

That seems more intricate.

[@shinoasada4590]

@ 2:35 what did you do, why cancel xln3 and 5ln2, what is the reason for it?

[@shinoasada4590]

Why did you not multiply the 32 on 2^x and 3^x

[@JohnBerry-q1h]

_no! no! no! no!_ NO… Instead, back at 1:22 , divide both sides by ln3 . This will create the term (ln2/ln3) on the left side of the '=' sign. Note that (ln2/ln3) = 0.630930 . We then will have... 0.630930( x - 5 ) = ( x + 1 ) 0.630930x - 3.154650 = x + 1 ...subtracting 0.630930x from both sides yields... -3.154650 = x - 0.630930x + 1 ...subtracting 1 from both sides yields... -3.154650 - 1 = x - 0.630930x -4.154650 = 0.36907x ...swapping sides yields... 0.369070x = -4.154650 ...solving for x yields... x = ( -4.154650/0.369070 ) *x = -11.257079*

[@canfriendly34]

I usually find X after W before Y🤭

[@deltalima6703]

Probably behind a space somewhere on a sign or a rocket or whatever.

[@dairoku2017]

I've been thinking (x-5)log2=(x+1)log3 (x-5)=(x+1)(log3/log2) (x-5)=(x+1)(1.585) (x-5)=1.585x+1.585 Gonna shortcut over here -0.585x=6.585 X= -11.257 Is this correct?

[@Kotaallen-qg7zr]

Yes, I also solved it like this Easiest way to solve this problem by log base 10 not natural log

[@madhurmurkiya9067]

Dont know why its irritating to see multiplying/dividing both sides and cancelling the terms instead of taking that term to the opposite side and cancelling terms🤡

[@OsamaBeenBallin_911]

Gore log hai bhai Ham Desi log ko solve krte dekhenge behosh pad jaenge 😂😂

[@typothetical]

When do you know when to use ln and when do you know when to use log?

[@AndersRisager]

At which grade would this be taught in your country?

[@kidnamedsolid3547]

Am I the only one who absolutely loves logs. It’s not that I find them really easy or anything, they’re just so awesome.

[@mind.journey]

I was thinking of replacing the 3 with 2^log(2,3) -> 2^(x-5) = 2^[log(2,3)×(x+1)] -> x - 5 = log(2,3) × (x+1) -> solve for x Is it valid?

[@ankitbhadra4105]

to think that I knew all those formulas you used and wrote on right side but still I didn't knew how putting them together will get me the answer. Thanks a lot. Any advice on how I can solve this thing of not knowing when to put and which things together to solve questions like this ?

[@themrhorseman]

I'm in 7th grade, so i tried to solve it like this. i know it looks bad and there might be some mistakes here and there, but what matters is i got to the right answer! 2^(x-5)=3^(x+1) 2^(x-5)=2^(log2(3^(x+1))) x-5=log2(3^(x+1)) x=log2(3^(x+1))+log2(2^5) x=log2(32*3^x*3) x=log2(96*3^x) log2(2^x)=log2(96*3^x) 2^x=96*3^x (2^x)/(3^x)=96 (2/3)^x=96 x=log2/3(96)

[@Relativemotion-dw2yb]

Bro youve studied log in 7th grade? When i was in 7th grade i was busy counting the leaves in my garden trees lol...btw good going

[@JustMe-pd8zm]

Even cooler is the fact that if you simplify the first answer they would be the same(I don't know if I'm right to be honest but iirc then that's cool) For example the numerator (ln 3 + 5 ln 2) can be rewritten as: ln 3 + ln 2⁵ or ln 3 +ln 32 Which is equal to ln (3•32) or ln 96 And the denominator can be rewritten as ln (2/3). Which means the equation can be written as [ ln 96/ln (2/3)] which is just equal to the second answer log base 2/3 of 96!

[@ain_li_shem_mekory]

This is very brilliant!

[@belaitedwaif2212]

You can solve this a lot quicker by just splitting up the exponents into 2^x, 2^-5, 3^x, and 3. Then isolating x is a matter of factoring it out of 3^x/2^x. Then you get log(1/(3*2^5)) with a logBASE of 3/2. The answer is -11.26

[@impastar5066]

you explained it so well and i could feel your excitement in solving for x

[@GaminghunterYTOfficial]

It's ≈-11.25

[@rutamupadhye1828]

We can also take log and rearrange the terms and then do divedendo.

[@theallmightycabbage]

I am still mesmerised by how smoothly he switches pens

[@MidoriMae]

I once decided i wasnt gonna go to uni after doing engineering math with intergration and differentiation. I did finish my diploma but pursued the arts afterwards. Looking at this video makes me want to reconsider the sciences which has always been my true interest. My understanding was always rock solid but the math always killed me since im bad at memorising and no teacher was good enough at explaing to a functional degree. If only i saw your videos back then.. things might be different now.

[@liamw.7937]

>plot both functions >see where they intersect >eyeball the approximate solution to one decimal place >call it a day and get a beer

[@shubhamsrivastava8213]

So simple . I did it in my head in like 10 seconds

[@Vasilis_Sky]

Awesome video, you explain everything so smoothly!!

[@epikbaconb9780]

This is probably gonna be useful in the future so thanks

[@LorxusIsAFox]

Not an approximation; the two answers are exactly the same. This we can see through a combination of the change of base formula and mainpulation of logs in the numerator and denominator of version #1: ln 3 + 5 ln 2 = ln 3 + ln 32 = ln 96; ln 2 - ln 3 = ln 2/3. As for a workable approximation, we should clean this up a bit first: for X = ln(96)/ln(2/3), we have X = -ln(96)/ln(3/2), and conveniently for log approximations, 3/2^2 ~ sqrt(5), 3/2^3 ~ sqrt(11). Now, 9870/9216 ~ 1.071, so we take 96^2 = 9216 ~ 9870 = 11^3 * sqrt(11) * sqrt(5) ~ (3/2)^23, 96 ~ (3/2)^23/2 for a final approximation of X = -11.5, which seems to agree OK with calculator results of ~-11.257.

[@CuriousMindDevesh]

I exactly do this and get a error of approx. 0.25

[@akchharasinghchauhan4525]

I have a doubt! At the step (x-5)ln2=(x+1)ln3 Can't we Directly substitute the value of log 2 and log 3 in the eqn?

[@whydoineedname4932]

Taking log on both sides we get (x-5)log2=(x+1)log3 now value of log 2 and log 3 with base e are 0.69 and 1.09 approx So 0.69x-3.45=1.09x+1.09 0.4x=-4.54 x=-4.54/0.4 which is -11.35 So the ans must be around -11.35!

[@wilsonfrye9987]

This is the smart way to do this problem. You have to find the log of both 2 and 3 in some base eventually, so do it immediately. With the common base, multiplication rule for exponents applies, then just distribute the log found to each exponent and solve the simple linear equation.

[@jb31842]

@3:25 If you squint you can also read it as "absolute value of n cubed plus five, times n squared..." 😛

[@canirunit8162]

I'm not sure why I'm clicking on this. I am an economics student and just reading about consumer behavior theory, and it contains Lagranian function, which I've never heard before and try to find wtf is that equation but anyway I'm satisfied with this video.

[@thomasjacobsen8277]

As a student going into my sophomore year next year I am quite happy that I understood all of this!

[@TheLobsterCopter5000]

I'm not sure which form of the result I prefer. The first form uses ln rather than a logarithm with an awkward base, but the second one looks neater.

[@thsxi]

Personally I did it by equating the bases. Let’s say so 3^x+1 became 2^(log2(3))(x+1) so then you can just do x - 5 = xlog2(3) + log2(3) and re-arrange for x. Ended up with x = (log2(3) + 5) / (1 - log2(3)) ≈ -11.257.

[@mrwgamer4556]

I wish i found this channel when i was in high school

[@nguyenhuuhiep1505]

Thanks you so much I have been struggle for the exact same question for a long time and now a week before my test I randomly see this video ❤

[@handlesarecringe957]

me, a researcher in aerospace engineering, watching this at 2 AM

[@janhulka1418]

i remember learing the first method in high school but our teacher used the common logorithem instead, was that incorrect or also possible?

[@EC4U2C_Studioz]

I prefer the method where you just use the appropriate log base from the start which leaves whatever is in the exponent.

[@mr_pc-sco197]

Watching this as a gcse student in the uk knowing this won’t come up in my exams but this was thoroughly interesting

[@anomalytm05]

Wait, log10's also okay to use for this, right?

[@kb-ly6dx]

This helped me so much, thank you!!!

[@rushilgupta1203]

I am from India and this is the most basic example that we solve in grade 11 🙃🫠

[@ISHTFAUA]

Here in brasil too lol

[@TheEulerID]

I have (5+log2(3))/(1-log2(3)) which I think produces the same value. There seems to be an allergy to using logs in bases other than e or 10 in official answers.

[@cristalmyth09]

I learned this a while ago, i kinda just forgot about it. So yeah, im greatful for the recap