Solve 2 to the x = 9 , what is x=? Can you solve this type of Algebra Equation?

[@thunder8cat]

I am 63 years old and never did well in math. I work around a bunch of engineers and am truly amazed with math at this point in my life. I love your videos. I like the way you explain simple math. I am at this level, so I really appreciate it.

[@cheriem432]

I find your comment very interesting. If I'm not being too personal, I am really curious to learn how you came to be working around a bunch of engineers without having done well with math? 🙂

[@thunder8cat]

@@cheriem432 In High School I worked a full day so my math was limited. I never went on to calculus and really didn't do great at algebra. I am now a construction manager for renewable power plants and work closely with all types of engineers.

[@warrenjohn]

Same for me. I am 52 years old and my best friends are engineers. Needless to say, I try not to discuss mathematics when they are around!

[@paulhandidjaja1579]

Me too I'm 68. He is good explaining this.

[@nancyconnolly2210]

Exactly. The appreciation of math is limitless, regardless of your level of experience with it. You’re not alone.

[@MarliesBraun]

We didn’t have calculators at my schools, we had logarithmic books and a slide rule. So I had to learn it the hard way, nobody would do this today! Thank you for this refresher math course!

[@phillipmcduffie9353]

I'm also 81, but I learned to use a calculator in 1973. I went to work for Texas Instruments. I only used a slide rule in high school chemistry and throughout my university years. Man, was I glad to only have to push a few buttons and look at a small display of red light segments. There was the answer !!!! I am speaking of the venerated TI SR 11 ( Slide Rule Replacement Calculator ). We were really having fun back in the old days. Our astronauts all carried slide rules to the moon and back. They didn't have calculators back then. I had to compete with senior Chemical Engineering students who had the $100 +-×÷ calculator available in the early 1970s. I was still using my slide rule. I didn't have a single dime to my name. I really had nothing back then. But, I did get an education. All worked out in the end.

[@dancingpixie6120]

I used a scientific calculator when everyone else had graphing calculators. I completely agree!

[@RedDogByrd]

I am 81 and I need to go find a slide rule. Anyway, I think you gave a great explanation. I looked at this in an effort to ward off dementia. I hope it works. You did a great job and keep up the work.

[@dazartingstall6680]

Interesting use of the phrase "round it off" there. log9 ÷ log2 = 3.169925... Rounding that off to three decimal places, we get 3.17. (The third place would be a nought, which we can omit.) 3.169 is a truncation, not a rounding.

[@jenohathazi920]

it's correct

[@dazartingstall6680]

@@jenohathazi920 I didn't say it's not correct. But it's a correct truncation, not, as John called it, a rounding off.

[@usic_imaging]

​@@dazartingstall6680 I think its more to avoid the arguments of when to use either method at the end of the vid as its a whole subject in itself . its a good point though !

[@Hugh-N-inKent]

Interesting.. not had to even think of logs for decades so a good revision for me. The phrase "round it off" there is very wrong. Using four figure log (tables) you can only quote the answer to three significant figures anyway but suppose you used six figure or were allowed to use a calculator then the question should either state or you should state as part of your answer the number of significant figures. So if you used a calculator say you might wish to quote to four figures giving us 3.170 and here the zero is a significant number. Extending the solution a little an astute student might point out that the rounding error is so infinitesimally small that for practical purposes the solution is 3.17

[@oahuhawaii2141]

The value to 3 decimal places is 3.170 . The way you wrote your value indicates 2 decimal places.

[@kennyhalltotga2272]

It took a little while to integrate this material into my understanding of mathematics, but with the help of this utube link I have grasped the concept. Thanks.

[@JohnBond-m4t]

Could you please differentiate the proof?

[@nancyholcombe8030]

I understood from basic knowledge of exponents that it would be 3-something. Thanks for explaining what a logarithm is!😊

[@robinhaas6895]

The log of "a" to the base "b", is equal to the log(a), divided by the log(b). Here, a = 9, and b = 2. Hence: x = log(9) / log(2)

[@michaelashall4523]

Thankyou for taking the time to fully explain the options to be considered.

[@TonyAllen1]

Without a calculator as sort of an approximation, I took 2+2+2+1 = 9, where 1 is 1/2 of 2 and split 3 ways is 1/3, so 1/3 of 1/2 is 1/6 or .16666 or .17 rounded to the hundreds, which is kinda close.

[@parthenogenesislove]

I didn't view the video. I can see why it would stump people. Fortunately I have an MS in Economics. We had to do this all the time in our econometrics classes. You take the natural log of both sides. ln(2^x) = ln(9). We can solve the right easily and it comes back to a number. With the left side when you take the natural log of something like that, you're just going to be left with the power, which in this case is x. So x=ln(9)=3.169. I got so used to the math I did in getting that MS, I went back and got a BS in maths as well. That degree definitely enlightened me and my way of thinking even though I work in a job that uses neither math nor economics really. I did subscribe and like the video though, because I think it is always great to promote maths.

[@petervladyevich8535]

2^x=9 exponential writing e^(x ln(2))=e^ln(9) x ln(2)=ln(9) x=ln(9)/ln(2) Thank you it's a pleasure to ear your explanations

[@Gnowop3]

Overcomplicated! X = log9(base2) by definition of logarithm

[@herrtio]

Please, for us (dummies) unfamiliar with these procedures can you explain why x squared gets added to one, and what happens with dx? A lot of the time it's these (perhaps insignificant) details left unexplained that leave many of us puzzled as to why a certain action is taken. Thank you.

[@tomtke7351]

2^x = 9 off top x=1 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 guess x = 3.2 for 2^x=9 this is an excellent problem for logs. log a = b or 10^b = a further loga^x = xloga [basic log rule] log2 = x where 10^x = 2 log2 = 0.3010 log9 = 0.9542 [found on calculator but back in my day we looked up these numbers in the back of our text book in log tables because we had no calculators. We even had logs of trig functions. The little black book for the engineer exam has (or ¿had?) these tables.] 10^0.3010=1.9999, =2 10^0.9542=8.9991, =9 here 2^x = 9 same on either side=> log { 2^x = 9 } log(2^x) = log(9) xlog2= log9 x= log9/log2 = 0.9542/0.3010 = 3.1699 our previous guess = 3.2 V E R I F Y (calculator) 2^3.1699 =? 9 8.9998 =❤ 9✔️

[@dazartingstall6680]

Further verification, again with a calculator: 2^(log9 ÷ log2) = 9 I also remember log tables. We had them in a thin book that also contained sine, cosine and tangent tables.

[@oahuhawaii2141]

You just committed an error with your calculator because you wrote .9542/.3010 = 3.169925 . You really used log(9)/log(2) , but wrote the truncated intermediate results of each log as you typed in the expression, and then cited more precision than what you wrote for the truncated log values. [With the full precision of the calculator, x ≈ 3.1699250014423...] If you actually entered .9542 and.3010 , you'll get 3.170{0996677740...} . This is good for 4 sig figs, 3.170 , because the inputs are only good for 4 sig figs. I simplified my solution using log(3): 2ˣ = 9 x = log(9)/log(2) = 2*log(3)/log(2) ≈ 2*0.47712/0.30103 ≈ 3.1699 I remember the log of 2 and 3 to many sig figs, and I used 5 places because their values drop off nicely at 0.30103 and 0.47712 . A simple 4-function calculator returns 3.1699{166196060...} , so the estimates for the logs yield 5 sig figs with 3.1699 .

[@mjsound12]

I love how you talk in circle so you can generate and log more minutes to KZbin 😅!

[@cooganalaska3249]

Too many words. Please get to the points faster.

[@harveybc]

I'm happy to say I remembered that Logs would be involved but in spite of remember doing this stuff I"d forgot the 2^x to x log 2 rewrite. How old you might ask. Well, after you showed how to do it I took out my slide rule and verified it. (Yes, I have one of those things on my computer desk. And I don't have to worrying about remembering to change the batteries 🤣)

[@imagseer]

I remember this from when we were taught about logs and slide rules! - Early 1960s. Thank you for stirring the memory.

[@tomtke7351]

add in: the slide rule is logarithmic laid out to essentislly ADD together two numbers' logs when multiplying them.

[@imagseer]

@@tomtke7351 It also used subtraction. I remember you had to mentally move the decimal point for bigger numbers. So you kind of needed the language of maths to use the higher functions. Maybe I'll dig it out and see how far I can still use it. My Dad had a circular one in a case.

[@newtonocharimenyenya2458]

❤❤❤.

[@marccracchiolo4935]

Great video it’s so easy to forget this stuff you make it easy to remember & understand

[@nancyconnolly2210]

It does start out as an algebraic equation. Logarithms is a strategy. You can also approximate it graphically and by substitution. Just focusing on logarithms limits a students willingness to strategize. I like their approach.

[@oahuhawaii2141]

2ˣ = 9 x = log(9)/log(2) = 2*log(3)/log(2) ≈ 2*0.47712/0.30103 ≈ 3.1699

[@carlosadrianob.blanco6521]

Determine if 2 to the square root of 2 plus 3 to the square root of 3 is a rational number. Exercise taken from an old math book I believe the author was Serge Lang.

[@rayrocher6887]

thanks good person. 1 tiny problem - it was a repeating decimal. but you are right. many problems - have good , simple , but excellent answers. in other words - try harder. keep the Faith. Math Good , amen.

[@MusengimanaEmmanuel-n8d]

You Man ,you spend a lot time to solve simple a question!!!

[@bambymed]

He is talking much too much really

[@nipamarkets]

But he was teaching bro I want to learn maths and I think he’s a great teacher for me

[@Drop_off_on_the_right]

Yet, if you went to high school or college, you sat through hours of lecture. You’re too smart and too cool for school. 😂

[@leofajardo713]

@@Drop_off_on_the_right does not make sense back then we hated math now we're older were fascinated by it.

[@srikrishnaahithanala3487]

It's for beginners / those who know first principals only.. I a BSc student of 1970s enjoyed the way..

[@StevenKoonse]

Just looked at it and decided it was around 3.2 w/o the rigmarole. As a pedestrian going through life, I have to constantly perform “it’s probably” calculations to get by,

[@paulords8327]

Detailed explanation made it more understanding for all knowing why they do what they do.

[@civilpse5458]

We used Clarks Logarithmic table at school and in college,it was slide rules. Really enjoyed reminiscing.

[@dennis2376]

This is one way of giving people headaches, but very cool. Thank you.

[@jessicaforman6774]

😂😂😂🎉

[@TimVeatch]

Once the definition of logarithm is shown at 6:30 what is there to show? 2^x=9 can be rewritten as log base 2 of 9 =x by definition. Then you can get the answer from the Google search bar QED No need to take the log of both sides and use a property or two. Then you can go into the story about how in the old days mathematians spent 20+ years calculating tables of logarithms.

[@latashawilliams6421]

I love watching different ways to solve equations. The only way I know to solve this question is xln2=ln9 x=ln9/In2. 😊

[@briankearns4771]

Holy crap, no wonder I fell asleep in math class.This is why everybody says math teachers are boring

[@nancyconnolly2210]

@@briankearns4771 Ha ha It all depends upon the teacher.

[@JeffVV9413]

Starting from 5:53 of the video, perhaps one more explanation, that 2 to the power 4 = 16. log 2 base 2 = 1, to explain log base 2 16 = 4. Bacon and egg is good to memorise, but how to understand is more than important.

[@NANDANKAUSHIK]

What if we don't have scientific calculator? I remember referring to anti-log tables to calculate log of a number to base 10 .

[@mikehigbee2320]

I have forgotten just about everything I ever learned about exponents and logarithms. I don't think I ever understood logs, anyway. You confused me by going in depth about the " bacon and eggs" leading me to see log base 2 of 9 equals x. But then you solved it by ignoring that and turning it into x times log (base 10) of 2 equals log9. So the REAL lesson for me was that log 2 to the x equals x times log 2. That was what I needed to know and you kind of just brushed that aside. Nevertheless, I (re)learned something, so thanks.

[@somtingwongwai7194]

Now go look at your student loan and see if waiting for someone to bring this topic up at dinner is worth it.

[@manuelteixeira2496]

i appreciate your kind effort in doing math explanation.

[@Lynn.Panadero4242]

It’s been years since I’ve done math like this. I remember looking up logarithms on the chart in out textbook. We didn’t have the cool calculators at that time.

[@russelllomando8460]

thanks for the lesson

[@isahisahk5459]

An experienced teacher.Thak you so much for making maths much easier.

[@DukeMitchell.223]

So how do you do the problem, all I learned was how to press log keys on a calculator. How do you get the difference from 8 to 9? Without a calculator

[@devonwilson5776]

Greetings. The identity can be solved by the use of log. Therefore, 2^X=9, 2^X=3^2, an Log2^x=Log3^2, XLog2=2Log3, and X=2Log3/Log2, X=2(1.58496)= 3.16993, X approximately equals 3.17.

[@oahuhawaii2141]

You're sloppy with your notation. x = 2*log(3)/log(2) x ≈ 2*1.58496 ≈ 3.16992 You shouldn't use "=" for an approximation. And once you cite a certain value for an approximation, you can't use another value with more precision. That is, the last digit is '2', not '3'. It's obvious you copied a few digits from your calculator, but continued the calculation with more precision. If you had cited x ≈ 2*1.584962500... , then you can show more digits than what you had typed.

[@moseschristiemukambo7730]

This is lovely. I love maths so much but unfortunately didn't reach this level of education. I would like to access your lessons to advance my knowledge and thanks for this ❤❤❤.

[@alreynolds4152]

I never got an explanation of Log before. Thank you

[@Gnowop3]

Where to start depends on what you want. In short, X is by definition log 9(base2)

[@santyclause8034]

Then he says common Log base 10, after a lengthy waffle on side-track.

[@rayrocher6887]

thanks for being and working ,trying to be a great Teacher. Good work.

[@ayoCanada0921]

Thanks for teaching this again at my age of over half a century. When I was going to school back then too young too understand this things. Like putting a kid in a rocket classroom.

[@gerlindes6624]

I love the reminders of 55 years ago, but I would like the presentation a little more direct. Whatever I will keep watching.

[@dagnt8145]

The morning after , this is just what I was looking for to get me through . We all could use some continuing education right about now .

[@mrb7094]

Just the same. This and cooking videos. And possibly a long trip somewhere....

[@BernardGreenberg]

The title of this should be "What is a logarithm?" not "Solve 2 to the x = 9". If you know what a logarithm is, this is no problem.

[@Leptospirosi]

And, if you want to know what a Log is. There are much better andnless confused explanation then this rambling

[@BernardGreenberg]

@@Leptospirosi Exactly.

[@PelosiStockPortfolio]

Why not reduce it to log2(9)? You wouldnt leave a final answer as 4/8 when you could write 1/2, right?

[@galtem]

I was really hoping this video would actually talk about how to calculate a Log; that's one area of algebra that's always been a mystery to me. I get that Log 9 is a value, but HOW do you calculate it? I was hopeful until you said to take out your calculator to find the final answer. I think that would go along way to helping folks understand Logs, if we knew what the calculator was actually doing instead of it being a magic box and spitting out some never-ending decimal number. I've just never been able to visualize what its doing. Thank you.

[@Steven-v6l]

Sorry, there is no simple calculation. Honest. To derive the formula used to calculate log(x), you need to use calculus. The basic calculation involves summing an infinite series - in practice keep adding successive terms until you have the accuracy you require. That's why most people use a calculator or a Big book of pre-calculated values of log(x). here is one formula for the natural log of x: ln(x) = sum (for k=1 to infinity) [ (1/k) * ((x-1)/x)^k ] valid for x >= ½ to get 15 decimal places of accuracy for ln(9) you need to evaluate about 250 terms of the form (1/k)*(8/9)^k . for k=1 to k=250 so ln(9) ~ 2.19722457733619 You need more digits, add up more terms.

[@PaulPlisiewicz]

@@Steven-v6l I shared that same interest in learning how to find the log; so thanks for sharing your knowledge in the reply. Your post makes a great argument for investing in a calculator!!

[@Kjetilstorm]

Man goes off on a tangent so much I get two adds between every section.

[@spunstricken9065]

The fun thing about getting older is that I can watch a movie and it is much like watching it for the first time. 😊. Same with Maths. I unused to be a Maths tutor in University and at a private tutorial center. I cat believe how much I have forgotten. Thank you.

[@johnrday2023]

Could have been fully explained in a tenth of Math Man's wordy complicated discussion !

[@canberradogfarts]

If you think he's "too wordy" then you're too not very bright.

[@kevingilliam6807]

I just rewrote the original equation as a log equation. log base 2 (9) = x. Which can be rewritten using rules of logs as Log 9/Log 2.

[@johnanthony4194]

This is a serious question. How many people NEED to know how to solve this equation? This leads to a further consideration and that is : In High School Maths already a lot of topics are included that 90% of pupils will NEVER need nor understand. This can mean that many pupils simply give up on ALL of Math thus leaving them struggling with daily calculations that are useful, and at risk of being cheated by people who can do basic math. I am not suggesting that a full math program should not be offered to those who have interest and ability. I have a daughter who is professor of astrophysics, but that is unusual. So she benefited from a comprehensive math course at school. On the other hand I recently looked at a Year 10 math course from England most of which I simply did not recognise. I concede that if I had understood calculus in 1961 my whole life would have been different, but not necessarily better.

[@jumperstartful]

Me and math are polar opposites and don't attract. I learned my times tables up to 12x12. And that served me very well for 60 years.

[@reddblackjack]

Yeah, I don't know what you're supposed to do, but logic suggests x=3.17 . 2³ is 8, and just a bit more is needed. But 2 to power of 3.2 is too much. 3.17 is actually a little too much, but close enough. I like it.

[@johnmarchington3146]

I looked up my Chambers seven decimal place log tables and found that log 9 (to base 10) was 0.9542 (to four decimal places) and log2 was 0.3010, so 0.9542/0.3010 = approx, 3.17 for x.

[@cedjulemckeever]

Just as a side, you could graph this in Desmos and get a pretty accurate answer.

[@AngieBurris-ry4hz]

What about doing the problem without the calculator?

[@johnopalko5223]

Use either a log table or a slide rule.

[@santyclause8034]

Those are also calculators. He means show your worked out solution using pen and paper. Did you even apply math?

[@druariel]

Log base 2 of 9 might be the preferred answer rather than the decimal approximation

[@rbono01]

So how do you calculate 2 to the 3.167 power?

[@neilduran3586]

2 x 2 x 2 x (2^0.167)

[@wilfredkube8570]

We were asked to find the value of x, in the equation 2^x = 9 Step 1: 2^x = 9 Step 2: x log2 = log9 Step 3: x = log9 / log2 Using a calculator, base10 logarithms (using the “log” button): Step 4: x = 0.954242509 / 0.301029996 Step 5: x = 3.169925001 Or: x = 3.1699 (to 4 decimal places) Or: x = 3.170 (rounded to 3 decimal places) If you wish to use natural logarithms (using the “ln” button), then: Step 3: x = log9 / log2 Step 4: x = 2.197224577 / 0.693147181 Step 5: x = 3.169925001 In other words, it does not matter whether you use base10 logarithms, or natural logarithms - just make sure that you are consistent and use the same base for each side of the equation. = = = Now, to find the answer (let us call it “a”) for “2 to the power of 3.169925001”, here are two methods which you could use to find the answer. Your scientific calculator is likely to have a button marked “log”. It should then also have an “antilog” function which is the inverse function for that “log” key, namely 10^x (10 to the power of x). The second method involves the y^x key, which I hope is present on your scientific calculator. Method 1: Step 1: a = 2^3.169925001 Step 2: log a = 3.169925001 * log 2 Step 3: log a = 3.169925001 * 0.301029996 Step 4: log a = 0.954242509 Step 5: antilog (log a) = antilog 0.954242509 Step 6: a = 9.000000000 Method 2: Step 1: a = 2^3.169925001 Step 2: Press the “2” key, then press the “y^x”, key, then enter 3.169925001, then press the “=” key. Step 3: the answer should display: 9.000000000

[@garoldcounts8777]

I'm intrested in this, I have worked in construction so math was needed, but I worked with shapes, in whar form of practice would this be used

[@Eligus33]

This would've been so helpful with my online intro to electrical engineering class

[@myobboy9973]

Mathematicians use the words "logarithm" and "exponential" far too carelessly. The log function takes an ordinary number and moves it into the logarithmic number domain. In the case log 10 =1 the 10 is a flat finger counting number and the 1 is the logarithm. But the logarithmic number is the same as the exponential number (base 10). Therefore the 16th century function name "logarithm" is redundant, we should rename it as, or just describe it as, the exponential transform function. This is important because we stop having two names for the 1 in my example. At present I can describe the 1 as a logarithmic number or as an exponential number but the 1 does not change, it is still the same 1 in the reverse exponential evaluation of the 1 i.e. 10^1 = 10. The point is that e^x should not be called the exponential function, it is the reverse exponential transform; the x is the natural exponential number and e^x decrypts the x into a flat finger counting number.

[@1234larry1]

This video is for students of fundamental algebra. Your comment, while interesting, belongs more on a blog of number theory. You wouldn’t walk into a first grade classroom, where they teach 1+1=2 and start explaining that without the additive property, mathematics has no concrete foundation. Their eyes would glaze over and they’d be shoving cookies in their mouths.

[@cristobalvalladares973]

You taught an old man something. Thank you!

[@TommyMorris-mn4nc]

Question: Do you provide these in different languages?

[@panlomito]

I must admit I do know this one has to be solved with log but I don't have the routine to get the equation right. Acoustics wasn't my thing at school. Edit: I couldn't stand I had forgotten these terrible log equations so I forced myself to learn this AND solving polynomials after dealing with 8^y + 2^y = 130 or (2^y)^3 + (2^y) = 130 with x = 2^y so x³ + x = 130 So first there is this 3th grade equation x³ + x -130 = 0 that I couldn't solve. But I learned to give a try with some factors of -130 being -/+ 1, 2, 5, 10 f(1) = 1³ + 1 - 130 neh f(2) = 2³ + 2 - 130 neh f(5) = 5³ + 5 - 130 = 0 BINGO so x = 5 now we have to find the quadratic equation (x - 5) . (x² + bx +26) 1 . x³ + 0 . x² + 1 . x - 130 x 5 5 25 130 -------------------------------------------------- + 1 5 26 0 so x² + 5x + 26 = 0 or (x - 5) . (x² + 5x + 26) = 0 D = b² - 4ac = 25 - 4 . 1 .26 = 25 - 104 = - 79 so no further rational solutions. x = 2^y = 5 and then the logs come in: y . log 2 = log 5 so y = log 5 / log 2 ~ 2.32 units

[@pennstatefan]

Take the natural log of both sides and one has xln2 =ln9. This is really xln2 = 2ln3. x = 2ln3/ln2.

[@pennstatefan]

x ~ 3.1699

[@RNMSC]

As a question, when I do lg9/log2 in my calculator, I get the value 3.16992500144 within the limits of my display. Wouldn't the 3 orders of magnitude precision give 3.170 instead of 3.169, as the next 9 is greater than 5, or is this a situation where because the number that one would round up from is an odd number, you don't round up? (At this level of precision, there isn't a hand tool you would use that would recognize the difference, though a machinist would probably complain.)

[@paiganesh3962]

You are a good teacher in Math

[@treejerk1]

You know, by inspection and knowing we can use logs of any base, we have x = log(2)9 = 3log(2)3

[@treejerk1]

Even with your answer, I would hope you would tell your students to reduce using log identities... notably use log a^b =b log a to get from log 9 to 3 log 3, then log a / log b = log(b)a to get log 9 / log 2 = log(2) 9 = 3 log(2) 3

[@MartinDuke]

Without a calculator, and per your first explanation wouldn't bet best answer be log2 (9) = x? I agree that using log10 (or ln) makes sense when using a calculator, but it adds a step of taking the log of both sides then doing the division, where as you can just use the identity.

[@mikesheth5370]

Simple! If you know log solution . Log of both side. and power can be multiplied X Log2 = Log8

[@raya.pawley3563]

Thank you!

[@colturn68]

If you took log to the base of 2, then the LHS would resolve to x. Then x=log base 2 of 9

[@jamesharmon4994]

This is exactly my answer.

[@TheSimCaptain]

The only problem is you won't find log to the base 2 in your calculator. You're still left with the same problem of finding the arithmetic value.

[@jamesharmon4994]

@@TheSimCaptain that's funny, I can do it on mine - it allows choosing your base.

[@TheSimCaptain]

@@jamesharmon4994 Not on most calculators. But you still need a calculator, as log base 2 of 9 isn't an arithmetic solution.

[@undercoveragent9889]

@@TheSimCaptain To convert Base 2 to base 10, using his B, A and E notation: '2^x=9' => 'log(base 2) 9 = x' => 'log A/log B = E' where 'A = 9', 'B = 2' and 'E' is the exponent and the value of 'x'/ :)

[@wlhgmk]

log both sides, move the X in front of the log2, divide log9 by log2. Answer solved. How about this one. 2 to the X power plus 8 to the power of x equals 130. Also invloves log but possibly not as you think.

[@ghasemsoleymani7976]

quite easy you can solve it using logarithmic solutions:x=log9/log2

[@ghulamahmad8954]

Very easy ! Just take log of both sides of the given equation and simply get solution. X=2log3/log2= 2*1.58496=3.169900025 This is the solution please

[@ericberman4193]

Good video. Thanks for producing.

[@williambunter3311]

Thank you for this sir,

[@AnilKumar-xl2te]

Let us assume x=1, then 2 9 x=2, then 4 9 x=3, then 8 9 x=4, then 16 9 x will be between 3 & 4 x is not a natural number 2^3 + 1^3 = 9

[@mrpunch72]

May I ask what application you use for the "chalkboard "?

[@johnrozzi687]

Should have solved it using log theorems. x = log (base 2) x 9, so x = 2log (base 2) x3, so x = 3, since 2log (base 2) =1.

[@sunnysharma5166]

Take log on both sides we get x=log9 to base 2

[@MAS1234P]

You don't have to take log on both sides to come up with the answer as it is by definition X=log9(base2). A lot of people struggle only because they forget about definitions

[@AlanOzzy2593]

Do you offer sooyside classes ?????

[@paulkurilecz4209]

This is how I would do it: 2^x = 9 ln(2^x) = ln(9) xln(2) = ln(9) x = ln(9)/ln(2)

[@WineSippingCowboy]

1 of my high school 🏫 STEM teachers (double major BS in Math and Physics a major liberal arts college in the East Coast) would give 1/2 credit. Why? He would point out that 9 = 3^2. Thus, (2 ln 3)/ ln 2 = x or 2 [(ln 3) / (ln 2)] as the final answer.

[@paulkurilecz4209]

@@WineSippingCowboy true, mine is a general solution whereas your solution only works if the right-hand side has a rational square root. How would you solve it if the right-hand side is, say, 11?

[@WineSippingCowboy]

@@paulkurilecz4209 Your solution would be the model. 11 in place of 9.

[@sylvieaubry1324]

Many thanks Blessing to you always

[@larsmothander9301]

Starts solving at 1:30

[@TheCarterKent]

"Alexa, 2 to the X power equals 9. Solve for X"

[@peteorengo5888]

You can solve this just by taking a few guesses and narrowing down from there. Doesn’t take a geniuos.

[@god_bika]

You missed out that log 9 based 10 turns in lg 9

[@taylormedia1000]

Spoken like a true teacher who spends a half an hour explaining "One" question, and running off on tangents

[@powerlink133]

Why isn't this the equation: Log (base2)9=X? I ask this based on the relationship you show at the 6:05 mark. I am just following your rule for 2(exp)X=9.

[@habalem5]

like the way you explain simple math

[@franhagaman6806]

Thanks your videos wake up this (almost ninety yearold brain every morning and are fun.

[@ibh9999]

Can we get the answer without the calculator ?

[@chbengtan7266]

Too much rubbish until I lost my patience

[@mrb7094]

Ok. What? I get 2 x 2 x 2 = 8, i.e. 2 raised to the power 3. I get that! And I get 2 raised to the power 4 is 16. I get that too! But what the heck is 2 raised to the power 3.169?? 2 x 2 x 2 x 0.169? Which is, of course, wrong. I realise that 0.169 is spread across each 2! So, dividing 0.169 by 3 (for each 2) and adding that to 2 it's very roughly something like (but isn't quite) 2.056333333 x 2.056333333 x 2.056333333 which sort of equals 9. I now understand what 2 raised to the power of 3.169 kind of is. But now you realise what you're dealing with here 😆