A Nice Olympiad Exponential Problem.

[@АрсенийМаркович-щ4з]

The logarithm of the number a on the base of b is by definition the number in which you need to raise b to get a. Therefore, when you saw that 2^x=5, you could immediately write an answer with a logarithm. No extra frills.

[@johnwhyck8545]

Shoot.. the problem attracted me to guess. If x was 2; then 68 was smaller than 130. If x was 3 then 520 was greater. So I guess around 2.5. 😂

[@boblobgobstopper13214]

@Repent and believe in Jesus Christ I don't think Jesus would've solved the equation for me

[@-rate6326]

​@repentandbelieveinJesusChrist2 i believe John was Asian dude

[@forexsanjose6090]

Yeah. Log a b is like a mark for “how many time of value a would get me the value b”.

[@maxshuford3482]

the extra steps are needed for a rigorous proof, if u need just the answer you would be correct

[@CrawliestCotter]

The trick to solve this one is to get to (2^x)^3+2^x=130 and notice that 130 is a decently small number for a polynomial of form y^3+y=130. Testing the first few integers will get you to 5^3+5 = 125+5 = 130. Which implies that 2^x = 5. Solve for x by taking log_2 of both sides and you get x=log_2(5).

[@topd3449]

Note that you need to prove that 2^x=5 is indeed the only solution. Best way to do that is to prove that f(x)=x³+x-130, x>0 is strictly increasing, which is obvious, but necessary. Saying this for the younger audience.

[@lesleyrobertson5465]

My 15yr old daughter worked it out faster. Using less stages.

[@hisoka2359]

@@lesleyrobertson5465not a competition but my 2yr old son did it in 5s using only 1 step

[@topd3449]

@@lesleyrobertson5465 Nice one, but we didn't ask :)

[@stefanodamilano]

​@@lesleyrobertson5465 Then show us

[@moeberry8226]

This is a horrible method, no one is going to guess to break up y into 25y and -26y. You could have just used rational root theorem or even vietas formulas to find the roots for y. And even when your checking the solution there is no need to approximate x for 2^x=5, because when you plug it back into the original equation you obtain 2^3logbase2(5) + 2^logbase2(5), which gives 130 exactly. Since the 3 gets moved back to the power to obtain 125 and then using definition of the inverse function. Since logbase2(x) is the inverse of 2^x.

[@flastkchance5312]

5% of 18-year-olds in Turkey do this

[@JossoJJossoJ]

To make it worse, Math Olympiad doesn't allow calculators, so there's absolutely no point to approximate the value of log_2 5

[@KRYPTOS_K5]

Vietas!

[@nashriqmazlan5181]

I thought im the only one feel this method was absurd 😅

[@lauroneto3360]

I don't dislike that method , but I didn't understand how he got it.

[@ChavoMysterio]

8^x+2^x=130 (2^x)^3+2^x-130=0 y^3+y-130=0 y^3-125+y-5=0 (y-5)(y^2+5y+25)+1(y-5)=0 (y-5)(y^2+5y+26)=0 y^2+5y+26=0 a=1, b=5, c=26 ☆=5^2-(4×1×26) ☆=25-104 ☆=-79 (solutions not possible) y-5=0 y=5 2^x=5 x=log_2(5) ❤🧡💛💚💙💜

[@Kushal21170]

I thought the very same

[@jcb3393]

Basically how I solved it as well.

[@ChavoMysterio]

@@jcb3393 using the difference of cubes theorem and factoring by grouping.

[@sanathsarma1087]

if we Substitute the xvalue in equation . The ans is 130.018 so wrong question. Sure

[@mbukiseniqwabe4881]

Atleast you can solve it

[@mathnetx]

when checking, no need to get value of log2,5. since 8^(log2,5) = (2^3)&(log2,5)= 2^log2,(5^3) = 5^3 = 125

[@jimg1928]

Just to summarize and clarify two comments already made, note that: log_b(b^a) = a because log and exponentiation are inverse functions when using the same base b, and this one fact can be used to both help solve and verify the answer as follows: 1) 2^x = 5 --> log_2(2^x) = log_2(5) --> x = log_2(5) which of course is the answer. 2) 8^x + 2^x = 2^3^x + 2^x = (2^x)^3 + 2^x = (5)^3 + 5 = 125 + 5 = 130 which of course verifies the answer.

[@sos3309]

Alles sehr willkürlich ...

[@janjansen3140]

​@Repent and believe in Jesus Christ fok off with your god. Only trouble religion has given us so far.

[@muwlgr]

great verification, much better than one in the original clip (with approximate numbers)

[@jozefpollak1865]

I have just realized that my life without this kind of math is so nice.

[@lesleyrobertson5465]

U r funny

[@y.kal08]

That's what I am literally studying Pain and suffering for a simple x²+x+1

[@pholdway5801]

Only just ?This leaves my head in such humility

[@qwang3118]

Let y = 2^x, then y^3 + y = 130. y(y^2+1) = 130 has only one real root. By the rational root theorem, the possible rational root is an integer, and it divides 130. 130 = 2*5*13. Trying these can easily determine that y = 5 is the solution. x = ln(5)/ln(2).

[@lioneatsthesheep8996]

I can't believe I watched this whole thing and nodded along as I told myself "I know exactly what this guy is talking about"

[@kalyanraman8566]

Really good videos,Thanks.Good luck

[@learncommunolizer]

Thanks, you too!

[@BurnsRubber]

Solved it numerically in my head. 1) No integer x satisfies equation. 2) The value of x must be between 2 and 3. 3) It’s likely closer to 2 than 3. 4) Evaluate a few cases (or perhaps just one) and find x is appropriately equal to 2.3. Not exact like log2(5) but certainly expedient.

[@agusludin7947]

In multiple choice question: A. 2.30 B. 2.31 C. 2.32 D. 2.33 Which one will you choose😁🤣

[@BurnsRubber]

@@agusludin7947 Four possibilities in the second decimal means at most two evaluations to compute the correct multiple choice answer.

[@curingd]

y^3 + y is a strictly increasing function for positive y and 5 is a solution to y^3 + y = 130, therefore it is the only valid one. No need for most of the algebra.

[@wmcomprev]

In the last step showing the alternate base, why not use log base 2 instead of log base 5. That would immediately give you x(1)=log base 2 of 5. It shortens the calculation.

[@zhishuihu6650]

这疑似在骗流量，把一个简单的题用二次替代法绕了个大圈子，130分拆也无必要，可以取个简单数，最后答案也不是个整数，无人能人工计算非特殊值的对数。

[@AFRO_KEEN]

Not good because that would immediately give you an answer of 2.32. 😅

[@AFRO_KEEN]

@Repent and believe in Jesus Christ All praises are to God for what we have.

[@rataflechera]

With y=2^x, we have the equation y³+y = 130, or y(y²+1) = 130. If this have an integer solution for y, then y | 130, y=5 seem appropriate, so let's check: 5²+1 = 26, 5×26 = 130, and this is an answer. x = log₂ 5. y³+y = 130 should have other two answers: y³+y-130 = (y-5)(y²+5y+26) = 0. The other answer: y = ½( -5 ±√{25-104} ) = ½( -5 ±i√79 ) = √26 e^{i something} there would be numerably infinite complex solutions for x, and x = log₂ 5 as the only real solution

[@douglasclockmaker8779]

I did maths at university for 3 years but have not used it for many years I enjoy your videos, takes me back Maths is fun , you have a passion so thanks and ignore the negative comments

[@DianeUgo]

I did as follows: 8x + 2x =130 10x (add 8 and 2 together) =130 130/10 =13 X =13 So... (8x13) + (2x13) =130 104 + 26 =130

[@TheAsaber]

Guys, this is olympiad. It is not just using a calculator. Please, do consider the effort he has made so that we can learn more.

[@loganshaw4527]

Well if it is more complex it could be 8^-2 + 2^7=130 or 8^2.25 + 2^1=130 if there are two different x values instead of the normal 1. It does not say solve for x. It asks how to solve.

[@lchan1977]

Wasnt good at maths when i was young, but i appreciate the different methods used to come to the final answer.

[@michelrocker9044]

Avec : X=2^x, (i.e: logX=xlog2),l'équation donnée peut s'écrire X³+X =5³+5,d'où X=5, log5=xlog2, x=log5/log2. (les amis, je vous laisse le soin du détail).Ceci pour dire qu'il faut en faire UN JEU plutôt qu'une laborieuse démarche scolaire . ▪︎observation ▪︎recherche d'astuces.

[@chanjacky5036]

wow you are so smart

[@danc.5509]

2 ^3X and 2^X is equal to 2^7 and 2^1 Not exact, but the right hand side and left hand side are all at base 2 A quick glance means X must be between 1 and 3, and (not precise) but 7 ÷ 3 is 2.33333

[@steadfasttinsoldier8659]

Thanks mate for interesting explaination. 8^(log 2 of 5) = (2^3)^(log 2 of 5) = (2^log 2 of 5)^ 3 = 5^ 3 = 125 No need to use approximate values.

[@learncommunolizer]

Thanks for that!

[@onetwo92]

Yeaaah. I have must check that solution is here.

[@abeonthehill166]

Very detailed explanation ! Thanks for sharing ……..

[@learncommunolizer]

You are welcome!

[@notsoancientpelican]

Attack intuitively. Let 2^x equal a, then a^3 + a = 130. By inspection, a = 5. Therefore x = log 5 / log 2 follows immediately. The rest is trivial calculation.

[@Dave-hv6dp]

This is how I was taught to solve cubic equations. Start off with some guesses. Didn't take long and you've explained it in 1/(26y-25y)🤷‍♂️ the time he tried to explain it

[@bigbrotherisasob]

Correct

[@eagleithrustx777]

I get that he’s showing a process (though it’s seems needlessly complicated) when (for this example at least) it’s so simple - the exponent for the 8 can only be 1 or 2. So take a guess and start with 2. 8x2=64 130-64=64 2x2x2x2x2x2=64 It’s like 7 seconds in your head - no paper required. Maybe an extra 5 seconds if you guess 8 exponent 1 and bring the 2 exponent up until it adds up to 130, but as soon as you see that 64x2=128 you realize that it must be 8 exponent 2 (cause it can’t be 1 or 3) plus 64 (which you already got on your way to 128). So you take one exponent off your 2 and bring your 8 exponent up to 2 to match your other 64 and you’ve got it. Why make something so simple so complicated?

[@cicik57]

solve t³ + t = 130 = 5*26; you see that t = 5 ( function is always accending), so x = log_2(5)

[@flastkchance5312]

👏👏

[@吳昶霖]

令A=2*x,則A的3次方=8*x 原式：A*3+A*1=130,則A（A*2+1）=130 A=5，則2*x=5,x=log5/log2 或x=log以2為底的5

[@ИринаТата-д2ф]

Большое спасибо. Очень интересно и познавательно.

[@MoonAney]

Amazing method being a math student these are very simple to understand

[@learncommunolizer]

Glad it was helpful! Thanks ❤️🙏❤️

[@hummit]

Watching this video reinforces my belief that I’m just not into math…..this kind of question never fails to give me headaches….

[@raregfx]

The equation suggests the value of X is not an integer and trying few numbers we realized the value of X should be a real number between 2 and 3. As the other side of the equation is an integer, the left side must be integer too. So, by evualuating the smallest term 2^x by x = 2 yields 4, the next integer is 5. So, the other term should be 125. If we equal each power to these integers accordingly: 8^x = 125 and 2^x = 5, applying logarithm rules and solving for x. The value is x = log(5)/log(2) or 2.3219... which turns to be equal to log(125)/log(8). I solved this by assuming 8^x and 2^x were both integers which in other cases might not be true.

[@adityaprakash2779]

Better method is after coming down to the equation y^3+y= 130, try to substitute nos. Y cannot be even, so put y= 1,3,5. y= 5 is a solution. Now, divide y^3+y-130 with y-5, and you get a quadratic equation which can easily be solved further. The put y= 2^x and here is the solution.

[@davidthomas2278]

I am useless at maths but even I spotted that he had a quadratic that even a cheap Casio will solve

[@avendreams2057]

you could even stop when u had 5 as an obvious solution. you need to make sure that's the only solution tho. Which it is, because we have an increasing function

[@slowmo6113]

Nice solution, I'll just tell my approach to this problem. Here' my 2 cents: 8^x+2^x=130 2^3x+2^x=130 Take 2^x=t for convenience t^3+t=130 t(t^2+1)=130 Factorizing 130 you will get, 130=2×5×13 Now just do hit and trial and u will get t=5 2^x=5 X=log5base2

130 is 1000010 in binary. Done.

[@niji_k]

Not a great solution in general; for this specific problem, magnificent.

[@TacoTitan48]

8^x + 2^x = 2^(3x) + 2^x = 2^x (2^(2x) +1) = 130. So we have product of two numbers that gives us 130, and one of them is the square of the other + 1. Candidates: 2, 65 -> 65 is not 2^2 +1 5, 26 -> 26 is 5^2 +1 10, 13 -> 13 is not 10^2 +1 Thus, 2^x = 5, thus x = log_2 (5). Of course this eliminates other answers in theory, because 130 could be a product of not only integers, but it's very quick.

[@QuynhNguyen-cw6px]

130=13.10=26.5=(25+1).5=5^3+5; T^3+T_130=0, (T=2^x >0); T^3-5^3+T-5=0; (T-5)(T^2+5T+5 +1)=0; vì (T^2+5T+6>0); T-5=0; T=5 =2^x;

[@Terruup]

I solved this in 20 seconds by just trying a few numbers. 8x13 = 104 2x13 = 26 26+104 =130 So "X" is 13. Pretty simple

[@ryanlevesque5370]

I had X at 13 . (8x13)+(2x13) = 130. So much for remedial math skills

[@jespersrensen3418]

Yes

[@yksh295]

Very interesting ❤

[@learncommunolizer]

Thanks and Welcome 🙏❤️🙏

[@walterwen2975]

A Nice Olympiad Exponential Problem: 8^x + 2^x = 130 8^x +2^x = 2^3x + 2^x = 130 = (5)(26) = (5)(25 + 1) = (5)(5^2 + 1) = 5^3 + 5 Convert the exponential base number 5 into 2 using logarithmic math: Let 2^n = 5, n = log5/log2 = 2.322; 5 = 2^2.322 8^x +2^x = 5^3 + 5 = (2^2.322)^3 + 2^2.322 = (2^3)^2.322 + 2^2.322 = 8^2.322 + 2^2.322; x = 2.322

[@walterwen2975]

Thanks 🙏

[@mrsulaman9901]

A very elegant solution. 👏

[@TeslaEdits_]

In the last part simply use logarithmic formula 8^(log5 base 2)+2^(log5 base 2) = 2^(log(5^3) base 2)+2^(log5 base 2)=5^3+5=130 .

[@allmightjunior6917]

Math is sexy when you know what you are doing, and terrifying when you don't know where to start

[@lesleyrobertson5465]

I agree sir

[@123MATEMATICAS123]

I RESOLVED SIMILARE IN MY IDIOM SPANISH. THANKS

[@JPTaquari]

If I may use my cell calculator, I find easy, X = 2,322 Prove: 125,018 + 5 = 130,018

[@KADblK]

Да, представить 130 в виде 26*5 с ходу не получится. Клёвая задачка.

[@darcybrummett7004]

This looks extremely complicated! But I sucked at college algebra and logarithmic formulas drove me crazy. I’m not one of those people who love math and find it fun. I look at all those log formulas and it seems like a foreign language. I just couldn’t get it.

[@arconeagain]

I can see now why the engineering of buildings, bridges and rollercoasters can go so terribly wrong.

[@alokekumarghosh2992]

There is a easy method for factorisation, which is called zero method. You go on putting y equal to 0,1,2,....andso on until the lhs vanishes. In this case y=5 makes lhs zero. Hence y-5 is a factor. Rest is easy.

[@furkannama3497]

Its very beautiful math problem

[@learncommunolizer]

Thanks and welcome!

[@ЕленаИвчина-э4т]

Почему после введения замены не применить деление на многочлен (у - 5)?? Очень долго шли к квадратному уравнению. И потом, что оно не имеет корней, видно уже из скобки - сумма всех положительных слагаемых не может быть = 0)) И очень, ну очень долго решаете показательное уравнение. Там 2 строчки по определению логарифма.

[@skantorizh]

За рубежом простые методы решения уравнений попросту не изучают. Нет необходимости. Именно поэтому типичная задача подготовленного выпускника школы отнесена к олимпиадным.

[@МатильдаКаземировна]

Сложно о простом.

[@manoelperezmartins3262]

O desenvolvimento da equação foi brilhante , porém no final ele complicou o resultado com tantas propriedades de log. Bastava calcular direto 2^x =5 ; x=ln5/ln2 ; x=2,3219281 .

[@satishagarwal9758]

You can also solve the problem by using remainder theorem.

[@tanyapunyo2074]

True, I don't know why he did so much hassle

[@itsaurinotariel]

a³+a=a(a²+1)=130 a²+1 might have 26 because of 130. a²=25 a=5 then we control the equation, it's true( what a surprise ). a is equal to 2*x. So x is equal to log2*5. That's simple

[@jim2376]

130 = 128 + 2 = 2^7 + 2^1 might be a good way to start.

[@TheDGomezzi]

I’m not sure I understand why that would help

[@jaimeduncan6167]

It’s nice how any computer programmer / engineer / scientist can say : it’s just 130 in binary and get an answer literally in 2 or 3 seconds. Now the fun part is searching for non integer solutions

[@София-о1ц]

Хорошо.Для тех кто не знает формулу" разница кубов."

[@muhammetislambedirbeyoglu3135]

Your writing is so beautiful 👏👏👏

[@learncommunolizer]

Oh thank you!

[@Marinar147]

Каждый ученик в моей школе мог решить эту задачу! Элементарно.

[@СергейБулкин-ъ1и]

Кавказская хвастливость?

[@ChristopherLester-gm1bj]

Spot that it's of form y^3+y=130, 125÷5=130, so y=5, undo the y, and you've solved it in your head. Unless you also want the Complex solution.

[@davidbrisbane7206]

Of course, there are an infinite number of complex solutions to this problem when non-principle branches are admitted.

[@k33pstrumming39]

That is why I say modern mathematics are bullshit!

[@behrensf84]

At Y^3 + Y = 130 I would take a few guesses at y. Try 4, gives 68 too low. Try 6 gives 222 too high 5 gives 130 so 2^x is 5

[@СветланаЗернова-г3э]

Так можно до ,,черной дыры" дорешаться...

[@chrisbuch6042]

One point I don't get. Why do you split up 130=26×5 ? Would this work with 130=13×10 or 130=65×2 as well? It feels like you would know the solution and then use it to solve the equation.

[@progredior0]

No, it doesn't work with other numbers. Based on the facts that there are other angry comments about this, and that he gave no explanation for it, I don't think that there's any reason behind it. He just split it that way because he knew that those numbers would work, and that's it. I just wasted a lot of time trying to come up with any other explanation... So, what is the moral? Never watch trash content like this again.

[@cystish]

While watching this video i had an urge to go back and continue college after 25 years dropping out 😥

[@yummygoy5138]

That's sad man

[@ikawba00]

Oh I still have no interest in going back to college. I also dropped out, and this video reminds me of why.

[@cystish]

@@yummygoy5138 The saddest part my mom lived the dream of me going back to finish college until she passed away 7 years ago. And that was the last reason that could make me actually consider going back.

[@Shirome7]

↓ *Japanese high school student's way* 8ᵡ+2ᵡ=130 y³+y-130=0 (y=2ᵡとおいた) y³-125+y-5=0 (y-5)(y²-5y+25)+(y-5)=0 (y-5)(y²-5y+26)=0 y²-5y+26=0の判別式をDとおくと D=(-5)²-4•1•26

[@АндрейТебякин-е4ф]

Ничего не понятно, но очень интересно !

[@dietermackenbach782]

А ещё говорят - математика - точная наука! Почему тогда ставится значок "поимерно"? Я решил этот пример без всяких log и х 8 в 10 степени + 2 в 20 стерени = 130. Это же надо так мозг людям запудрить!? 5 лет в школе изучали алгебру, в жизни так нигде и не пригодилассь в отличии от геометрии!

[@marciliopc]

when you have y^3+y=130 the problem is finished since 130 =125+5=5^3+5 and y=5

[@IrinaI5]

Как я любила решать подобные примеры в школе!

[@helinamensah251]

💯

[@ПростоЦарь-й1м]

Кошматерный сон для меня))), зато зиготы, гаметы и прочие апендициты👍, одуванчики с пестиками и амёбы🥰!

[@seanthorson4542]

The best way to make kids hate calculus! Bravo 👏

[@Prem-K007]

Where TF ... Did he use calculus ? It's simple algebraic factorisation and simplifications including logarithms.

[@BigMC-J]

I really believe in the trial and error method for this one. It literally took me 90 seconds. X=2 was to low. X=3 was too high. A few shots in the middle and after 5 tries X= 2.321. Sometimes it's more efficient to just use logic.

[@StCreed]

Which would give you zero points on any math exam I ever took.

[@BigMC-J]

@@StCreed You never took a multiple choice test then huh.

[@ChasOnErie]

That is not a proof !!!!

[@BigMC-J]

@@ChasOnErieI didn't see that it was a proof. I seen an equation, I solved it.

[@kmj1976]

This is for an Olympiad, who’s to say you didn’t look over at someone’s work to get the answer…. that said, your approach would work just nicely in a game show

[@sunny0042]

you are very good at explaining things

[@learncommunolizer]

Thanks and Welcome!

[@lubosdostal8523]

How can I apply this method to a similar equation 8^x + 2^x = 131? Or 8^x + 2^x = 132?

[@ashleyhubbard8854]

For the general case you would reduce it to the cubic in y, as per the video. You could then try Cardano's method to solve the cubic.

[@lubosdostal8523]

@@ashleyhubbard8854 Exatly. The above shown way only works, if you have known/guessed a solution beforehand. What I wanted to say is, that the method works only for RHS 130 and not any others.. It is not generally appliable... You can solve the problem: 1) to guess a solution and reduce to the quadratic equation. 2) to apply rational root theorem, and try all divisors of 130, if they are a solution. I.e. y = +-1,+-2,+-5,,+-10,+-13,+-26,+-65,+-130. And again reduce. 3) to use a variant of cardano formula 4) to use a numerical software ... Or to ask WolframAlpha "reduce y^3+y-130".

[@ashleyhubbard8854]

@@lubosdostal8523 You are right. The 130 is obviously a contrived number.

[@georgesadler7830]

Thank you for the video.

[@learncommunolizer]

You're welcome

[@seanmaher3518]

This is why I majored in history 😂

[@your_virtuoso]

Very interesting method I would have tried it differently but thata the beauty of math. There's always another way

[@learncommunolizer]

Glad you liked it

[@Бендекс-ю6ф]

ОХРЕНЕТЬ !!!

[@gurkanorhan9003]

For me the easiest method is adding together the 8 and the 2 then you get 10 . 13 divided by 10 is 13 and you get the answer

[@mrstudent9125]

Cool

[@learncommunolizer]

Thanks 🙏❤️🙏

[@sdgsdg9534]

This is an exercise that just does not need to undertaken. Go enjoy life.

[@yourownsde]

Valu

[@Maths302]

Very good

[@learncommunolizer]

Thanks

[@СергейГромов-ж7ф]

It is very good it is excellent

[@seanychenav]

I noticed that the verification at the end could be exact: 2^(log_base2 of 5) = 5 exactly. Likewise, 8^(log_base2 of 5) = [2^(log_base2 of 5)]^3 = 5^3. Hence, 5^3 + 5 = 130.

[@sk_jarale]

Excellent

[@learncommunolizer]

Thank you! Cheers!

[@samig9032]

The answer to y^3+y=130 is obviously y=5 by inspection. If you want to make these a challenge, you have to pick problems that don't involve small, whole numbers...

[@АнтонМ-з3е]

Спасибо, отличное решение для неправильного уравнения!

[@nadertalebi2365]

It would be faster and easer by logarithmic means .

[@omprakshgurrapu5063]

Just apply log both side logx= log130/4, x=log-2(5)

[@davidwoodhead1037]

Wow 🤯 amazing work

[@learncommunolizer]

Thank you very much!

[@skantorizh]

M-m, nice problem. I'll give it to my school students. It may be solved just in mind, with some experience.

[@TheVoitel]

The problem with the solution (or rather the problem itself) is that log_2(5) _might_ look like an explicit solution when in fact it is not exactly. Because log_2(5) means by definition "the solution to 2^x = 5". So this solution is in fact not really that much more satisfactory than "the solution to 2^x + 2^(3x) = 130". It is remarkable how satisfied we can get solving something exactly by reducing it to something else we can only solve numerically ...

[@Scien_Tific]

I don't really understand your point. Yes, x = log_2(5) is mathematically equivalent to 8^x + 2^x = 130, but what part of it makes it a non-exact solution? Are you saying it's not exact because it's not a number? Afaik log_2(5) can't be numerically represented (in decimal anyway), so log_2(5) _is_ the most exact you'll get. Really, the whole process of "solving" equations is a little misleading since the word implies you're discovering something, but really you're just rearranging things in a way that produces the most boiled-down, generic result that is (hopefully) more human readable than the gibberish you started with. Your implication of numeric values being more correct than expressions like sqrt(2) is imho kind of strange. If you need to represent log_2(5) as a numeric value for whatever reason, you can always calculate it and, depending on the application, choose the best accuracy to go with, but none of those values will ever be equivalent to log_2(5), precisely.

[@TheVoitel]

@@Scien_Tific *Solving an equation* usually means to find explicit forms for the solutions of an equation. And my point is that while x = log_2(5) may look like an reasonably explicit form behind the bars it is in fact defined implicitely (like many function inverses). If in some theory we’d need such things a lot we might define a function such as F(k, y) as the solution to 2^x + 2^(kx) - y = 0. Then suddenly the solution of this equation could also simply be x = F(3, 130), which is by no means more complicated than x = log_2(5) (apart from being more esoteric). This is something quite common in mathematics, and there is nothing wrong with it. But point I’m making here is that an exercise *solve this equation* by itself is very vague as long as we do not know what is considered a solution. Because if anything that by itself is defined as solution to an equation doing this becomes quite trivial.

[@Scien_Tific]

@@TheVoitel Fair enough, I see what you mean, although one could argue that in a school setting most people would probably have a pretty decent idea of what constitutes to a solution, other than a few possible edge cases. But I'm also not an expert in these things by any stretch of the imagination so really I don't feel qualified (nor compelled for that matter) to oppose what you're saying.

[@DineshSingh-xv3bu]

So hard. So Nicely explained

[@surajupadhyay8847]

Great sir

[@АндрейКопаев-ш2р]

Посчитал сам и проверил. Ответ неверный! Почти на 0.5! К тому же здесь обещали РАВЕНСТВО, а не примерный ответ. То ли задача поставлена неверно, то ли решение в корне неправильно.

[@heliomaster125]

Это решается с помощью теоремы Безу. Сначала замена переменной, а потом сама теорема.

[@irkhrx03]

Please tell me how it helps and works in our daily life ?

[@60SecondsWithKishoreKumar]

Good method when you have to solve only one question in 3 hours...

[@janholcik9715]

substitution : y= 2^x; then: y^3+y = 130 = 125 +5 = 5^3 + 5 => y = 5; back to substitution: x = (ln5/ln2)

[@josephmalone253]

This video and comments was helpful thank you.

[@learncommunolizer]

Glad it was helpful!

[@andreasraab5294]

Ich komme mir nun freilich blöd vor, zu bemerken, dass ich mir blöd vorgekommen war, als ich sah, dass es ja nun doch etwas gedauert hatte, ehe ich gesehen hatte, dass das ja ganz einfach war und nun so schön, wenngleich nicht rational, aufgeht. 11 Minuten hat das aber nicht gedauert. Ein recht hübsches Molekül. Ich danke.

[@mathematicshobby5519]

I like your solution