A Nice Olympiad Exponential Problem

[@蔡秉杰-x8i]

you don't have to do the numerical calculation to check. They are all approximations and inaccurate. log(((1+sqrt(5))/2) / log (3/2) is a good enough final answer.

[@margarita8442]

I need to make a log

[@serkan5951]

@@margarita8442 what is log i heard that a lot in some kind of movies. "Log logg log log mmh "

[@zhartaunik]

The same thoughts. If you decided to calculate you could do it immediately at 7:53 But as said the thread starter better to leave the answer from 7:53

[@alec734]

@@serkan5951 what is log? Baby don't hurt me, don't hurt me, no more

[@benjaminmatte5225]

2

[@pangeo8183]

The value (1+sqrt(5)) /2 is called the golden ratio and is a really important quantity in maths. So we can write the solution as log_3/2(phi) where phi is the golden ratio.

[@thundercraft0496]

Same thoughts

[@hb1338]

The golden ratio is also very important in other fields, such as architecture and music.

[@avendreams2057]

φ is the letter you are looking for

[@harrisonrower2293]

I couldn’t believe seeing phi inside the solution when I did this one. Took me a bit to get through the algebra. Glad other people noticed too

[@syzrael1]

Please explain more

[@lanceschaina3084]

I rewrote as 9^x - 6^x - 4^x = 0, redefined A = 3^x, B = 2^x, formed the quadratic A^2 - AB - B^2 = 0, used quadratic formula, got A = [(1+sqrt(5))/2]*B, substitute for A and B, 3^x = [(1+sqrt(5))/2]*2^x, divided both sides by 2^x, got (3/2)^x = [(1+sqrt(5))/2], took ln both sides, solved for x. Took me about 90 seconds. I'm sorry, but, wow, what an ugly problem. Math Olympiad was really scraping the bottom of the barrel for this one.

[@arifayeasmin600]

math olympiad has catagories remember?

[@Vermiliontea]

That is a very elegant solution 👍 . However, it requires you to "see" things (even if it wasn't terribly obscure). Learncommunolizer's solution is a methodical isolation of X, which is why I appreciate it more.

[@ashamenai1659]

Yes I did similar but I completed the square instead of using quadratic formula.

[@infinitexgg]

Wrong, you cant do a second grade equation if a is not by multiplied by X²

[@misterroboto1]

@@ashamenai1659 The quadratic formula is an application of completing the square ;)

[@감각적으로직관이-j4w]

my solution is this. Let 3^x is A, 2^x is B A^2-AB-B^2=0 A=(1+sqrt5)B/2 because A/B is (3/2)^x, x is log(1+sqrt5/2)/log3/2

[@60fps-games]

To solve the equation 9^x - 6^x = 4^x, we can rearrange it as follows: 9^x = 6^x + 4^x Then, we can use logarithms to solve the equation: log9(9^x) = log9(6^x + 4^x) x = log9(6^x + 4^x) To solve this equation, we can use a simple iterative method. Start by choosing an arbitrary value x0 and apply the following formula until the desired accuracy is achieved: xn+1 = log9(6^xn + 4^xn) After a few iterations, the value of xn will converge to the solution of the equation.

[@MegoElazab]

You approximate 3times You have to approximate o će finaly

[@士雪]

A very nice piece of demonstration of how to solve for x. However, it is unwise to make approximation at the end part of the demonstration. Leave x as an irrational number is good enough.

[@marcosreal11]

It helps to discuss the strategy first. How do we know to divide by 4^x? It is from the underlying strategy, try to write all exponential terms with a common base. The wise student will make a reference sheet of the different strategies that can be used for different problems, since often the most difficult part is knowing how to start the problem.

[@davidzhang2319]

What would happen if you couldnt? What would you do if on the right hand side it had 5^x? You wouldn't be able to find a common base because 9 and 6 dont divide into 5.

[@ostasvl4203]

Try to divide by 6^x, or by 9^x. They're all the same

[@kimutaiboit8516]

​@davidzhang2319 you will require calculus to do that. As for this problem you can divide by any. I divided by 6

[@snowlynx5221]

Listening this solution, I forgot the math I knew, but I learned Indian pronunciation. Thank you so much!!!

[@rubensramos6458]

If a problem can be solved by using the Lambert W function, then it is certainty an exponential problem. On the other hand, if the problem can be solved by using the Lamabert-Tsallis Wq function (as it happens for a^x-b^x=c^x) then it is a polynomial problem.

[@Ez55upgogo]

12:59

[@FractAlkemist]

I wrote a simple Genetic Algorithm in Python for equations like this, which I frequently play around with. The GA does not rigorously solve the equation algebraically, but it solves it numerically and gives an answer as close to exact as you want - good enough for engineers, if not theoreticians. My program ran for about 3 seconds and gave a value for x: x = 1.1868143902809818 . With that value, the left side of the equation evaluates to 5.182430200563065 , and the right to 5.182430200563064. The difference is then 8.881784197001252e-16 , essentially zero; equation solved. .

[@thetimeiable]

Using code is too overpowered

[@FractAlkemist]

@@thetimeiable coding rocks!

[@FarQuZeDesigns]

Man my math skills are really rusty, but it was super awesome to watch this you literally move through it step by step, i like that :) Wish you all the best!

[@learncommunolizer]

Glad you enjoyed it!

[@PeterSedesse]

I was doing fine until the log showed up

[@dDAMKErkk]

[@mksarav75]

great and simple teaching. thanks.

[@learncommunolizer]

You are welcome! 🙏❤️🙏

[@sncarlos1970]

By using Euler equation Exp(i*pi)=-1 or e^(i*3.1416)=-1 you can get a complex result for any negative argument of logarithmic. So ln[1-sqr(5)}= ln [-1*(sqr(5)-1)]=ln(-1)+ln[sqr(5)-1) =ln [Exp(i*pi)]+ln [sqr(5)-1] = (0.211935,3.1416)

[@iwillkechu]

Way easier answer……… A= 2021, B= 0, C= 2020. Plug that in for both equations and it works for both.

[@nagyzoli]

Log turned into decimal is approximation, you loose accuracy(very simple.. can you get that value without calculator or engineering tables? Probably not. Hence.. approximation). The correct mathematical solution (at least the one that would have been excepted in my country) is the log form. log(3/2) ( (1+sqrt(5))/2)

[@kenanerol6037]

Exactly

[@TheMathArt]

watch this nice math problem : kzbin.info/www/bejne/aaLFpYdrZaaaZs0

[@42_universe]

indeed!

[@nikola2126]

Very nice congratulations👏👏👏

[@learncommunolizer]

🤝👍👍

[@EmmanuelBrandt]

very interesting approach, many thank for this problem

[@learncommunolizer]

Thanks and Welcome 🙏❤️🙏

[@michaeldolaberidze670]

Thanks 😊... This is Abracadabra 😊🌟💌👈9™-6™=3™...TM=1... This is True.

[@michaeldolaberidze670]

Thanks 😊...

[@learncommunolizer]

You're welcome 😊

[@jodhvirsingh8677]

Exponentials...quadratics...logs....Such a simple looking question and yet amazing

[@math_qz_2]

Very instructive task

[@johnfernau6299]

Great explanation! Towards the end of your video, I think I heard a cat purring. 🤣

[@learncommunolizer]

😁 You are right !

[@mohamedhasan7309]

Excellent question, and good explanation, thanks.

[@learncommunolizer]

You are welcome!🙏❤️

[@SaiSuhasiniRamalingam-n5s]

Really nice 👌

[@JoySaha-jf7oq]

Thank make more videos. Good luck 👍❤❤

[@learncommunolizer]

Thank you, I will try my best 👍❤️❤️❤️ @Joysaha

[@haiyangwan2363]

The BBC BASIC computing coding below can also obtain a answer by way of an iterative process. The coding requires you to state how precise an answer you want. At the level of precision in the coding below (determined by the line "UNTIL D

[@ideegeniali]

I didn't expect to see BASIC in 2023! I was born in 1978 learned basic on commodore 64, then went to Pascal, then to C, then Java, then Python. Long live Basic!

[@haiyangwan2363]

@@ideegeniali Thanks for your comment. I first learnt computing in the early 1980s using BASIC, then did a bit of Pascal & Fortran, machine code at University, then did no computing for 30 years. I bought BBC BASIC for Windows in 2017 and it seems to do all I want it to do. Subsequently i tried to get into Python but it was not immediatly obvious how Python was better than BASIS for my purposes. I suspect Python probably is a better but having just re-learnt BASIC I could not get into it.

[@davidfriedlander5512]

Very impressive. Nice work.

[@learncommunolizer]

Thank you! Cheers!

[@swindler1959]

my man just solved creation of universe

[@rogerfisken4189]

I don't understand the need to take logs to base 3/2 (at 6.25). Log calculations taken to base 10 or ln will work just as well and save a lot of time. You can just say xlog (3/2) = log ((1 + rt 5)/2) and go from there. The fewer steps taken, the lower the risk of errors.

[@ragingxavier2691]

Bruh, I'll be extremely honest with you. I was forced to take additional math as part of my curriculum in college, majoring in sports coaching. Why? I have 0 clue. This video brought about nostalgia but a rather hated topic in math that I had to learn. I only scored some points for the question cuz I couldn't solve the rest of it. Miraculously, I somehow scored a B for Additional Math XD. I really respect the fact you and everyone who understands and are able to apply and explain this. Hats off to you and everyone else, mate.

[@learncommunolizer]

Thanks 🙏❤️

[@viswathkkumar8334]

Bringing back my school memories, we use to do these in 6 to 8 th grades

[@justarandomguy77]

Fekuu 😆😆

[@greninja7619]

Kya fek raha hain bhai,jee mains se tougher sawaal hai yeh sab

[@justarandomguy77]

@@greninja7619 nhi hai shayd tumne jee mains ke questions nhi dekhe

[@greninja7619]

@@justarandomguy77 lol jee mains me itne aasan sawaal aane lage hai bhai especially coordinate geometry ke, yehnbhi aasan tha lekin jee mains tough nahi hai, jee advanced tough hai

[@justarandomguy77]

@@greninja7619 9^x - 6^x = 4^x yeh question or If log2 (9 2𝛼−4 + 13) −log2 (3 2𝛼−4 . 5 2 + 1) = 2, then maximum integral value of 𝛽 for which equation, 𝑥 2 − ((∑ 𝛼) 2𝑥)+ ∑(𝛼 +1) 2𝛽 = 0 has real roots is ______. yeh question mai tujhe fark nhi dikh rha? Kaha jee mains ka question kaha yeh 😶

[@cielblue235]

a^2 - ab - b^2 = 0 where a = 3^x, b = 2^x with a, b > 0. Thus, u^2 - u - 1 = 0, u = (a/b) ^x > 0, u = (1+\/ 5)/2, yielding x = Log_ (p) (1+\/ 5)/2 with p = 3/2.

[@mikesteele5935]

This is a nice problem that illustrates a very general principle: "exponential problems" are usually solved by substitutions that transform them into non-exponential problems. The usual candidate is a substituion that gives us a quadratic equation. There are usually two or three feasible transformations, so not much creative work is required. One then gets some good practice with the more routine bits of school algebra. Several examples on this channel illustrate this principle.

[@TheMathArt]

watch this nice logarithmic problem : kzbin.info/www/bejne/aaLFpYdrZaaaZs0

[@magicman-sv3jm]

9x - 6x = 3x So now our equation becomes: 3x = 4x To solve for x, we can subtract 3x from both sides of the equation: 3x - 3x = 4x - 3x Simplifying the right-hand side: x = 0 Therefore, x = 0.

[@arnavkmr3895]

This magic man really did do something magical here! So many mathematical atrocities! Elegant!

[@magicman-sv3jm]

@@arnavkmr3895 It's was a joke ;)

[@arnavkmr3895]

@@magicman-sv3jm and clearly a funny and thought out one :) and actually everything is accurate if we drop the x from exponents to product, so you've been charged free from your atrocities magic man... for now

[@wowcheers7488]

I do admire and feel the sincerity of your content. More success and support to you. Cheers. 💕

[@learncommunolizer]

Thank you so much 🤗 @wowcheers7488

[@Goldieridesbike]

Cool video!

[@learncommunolizer]

Thanks! Glad you enjoyed it

[@supersbbrawl4ever]

Reminds me of when I was 16... All those log rules really fell straight out of my brain, or so I thought. Stuck within the deep recesses of my memory, they were residing in this whole time.

[@lordndrew]

Yeah. I’m surprised I Rem the log rules even though I learnt them 23 years ago.

[@supersbbrawl4ever]

@@lordndrew Wow, 23 years and its still all there too? Nine years in my case

[@studycat2658]

Great explanation! Very interesting problem

[@learncommunolizer]

Glad it was helpful!

[@NeelBasu]

Watching that without sound. Looked like a satisfying video.

[@learncommunolizer]

Glad you enjoyed it!

[@charlesmitchell5841]

Very good explanation going through all the steps. Thanks! 👍

[@learncommunolizer]

You're welcome!👍👍👍

[@Yeutoan-88]

Good

[@masatenisiTR]

Math skills refreshed, thanx...

[@soumyshrivastav5561]

Nice que. Thanks bro

[@learncommunolizer]

Thanks too! Your Welcome!

[@AngeloMastroberardino]

Very nice explanation

[@learncommunolizer]

Thanks and welcome

[@AprendiendoAprenderFerrero]

Good explanation

[@learncommunolizer]

Thanks and welcome

[@vladimirrainish841]

Let's divide both parts on 6ˣ. We get (3/2)ˣ - 1 = (2/3)ˣ. If y = (3/2)ˣ then y - 1 = 1/y y² · y - 1 = 0 so y = (1 + √5)/2 hence x = log₃/₂(1+√5)/2

[@jim2376]

Problems like this are essentially formulaic. So ln(golden ratio)/ln(3/2).

[@kwstaskalaboukas8306]

Beautiful content bro,but this ain't Olympiad material✌️♥️

[@learncommunolizer]

Yes it is

[@alainbel64]

​@@learncommunolizer q@

[@alainbel64]

QAa​@@learncommunolizer aaa

[@alainbel64]

Aa11

[@nagyzoli]

@@learncommunolizer To be fair nope.. at least in Romania, this is high school level 10th grade problem. Olimpiads would use an example that does not simplify to biquadratic and you would employ Taylor series derivations

[@JSSTyger]

Very good content.

[@learncommunolizer]

Much appreciated & Glad you liked it!

[@mbzumzum500]

Vrlo konplikovano,veoma nejasno objašnjenj

[@GOUTAM-wi3wi]

Brilliant. . . . . . . Excellent. . . . . . . . Nonsense ॥

[@DillVideos]

well done! good jo!

[@learncommunolizer]

Thank you! 😃

[@Anubis10110]

It was refreshing...love your accent

[@VagishaDas]

It's around 1.2 I used geometrical solution. Set on a graph both side of the equation using different options with whole numbers. Where the graphs meet you find the solution.

[@sdias1]

Being a post graduate in maths i felt the maths difficulty level first time in my life after seeing your pace and even writing minute details :- maths is dead now

[@tonitalas1757]

Beautiful problem, but not tough enough for Olympiad

[@karlaescorza1461]

9z

[@karlaescorza1461]

Zz

[@karlaescorza1461]

z😊si 😊z

[@karlaescorza1461]

00p😅

[@benben5861]

I like the way you present it, Bravo!!

[@MarcinBarylski]

Asrrf

[@MarcinBarylski]

Hjp😅 Jgj

[@pauljdowney]

humbly... rewrote as.. 2^x*1.5^x*2^x*1.5^x-2^x*1.5^x*2^x-2^x*2^x = 0 1.5^x*1.5^x - 1.5^x - 1 =0 define u = 1.5^x u^2 -u - 1 =0 u = (1+- sqrt(5))/2 = 1.5^x take the ln of both sides x = ln(1+sqrt(5)) / ln(1.5) = 1.18681439 approx , drop the negative (ln of negative)

[@jcherrera70]

Great job. I would consider the problem solved at 8:41

[@555cr]

The difference between the logarithmic calculation and the Newton-Raphson method lies in the nature and approach of the solution. In calculating with logarithms, I applied logarithms to both sides of the equation and tried to simplify the equation by bringing the exponent forward. However, this approach works only if the equations are simplified by their logarithmic properties. The correct solution was not obtained because the given equation was not in such a simple form. The Newton-Raphson method, on the other hand, is a numerical technique that uses iterative computations to find approximate solutions to equations. This method allows you to iteratively approach the solution using the derivative of the function. Starting from an initial value, iteratively computes an approximate solution, repeating the iterations until the convergence condition is met. The results of the Newton-Raphson method may vary depending on how the initial values ​​are selected and the convergence conditions are set. Also, the amount of computation and the speed of convergence may vary from problem to problem. As mentioned above, the difference between the calculation using logarithms and the Newton-Raphson method is due to the difference in the nature and approach of the solution method. Calculations using logarithms are limited to simple forms, while the Newton-Raphson method is used to find approximate solutions to general equations.

[@yasarkosan7364]

What a long and beutiful journey

[@bernym4047]

Impressive and very clearly explained. I was not aware of the several rules that you employed to transform the formulae si I need to research those to understand this in full. But thank you for demonstrating.

[@learncommunolizer]

Thanks and Welcome ✌️

[@ceccavara]

i like how most kept on watching without having a clue of what is going on, like me.

[@nikhilsmathematics7048]

Very nice explanation. Because of this video I have revised 3 chapters, quadratic equations, logarithms, exponents

[@learncommunolizer]

Excellent.All the best!

[@f910104]

if you approximate to decimal during Olympiad, your answer is wrong. In addition, if approximation is allowed, simply using a brute force with the expected precision with a small piece of code will do better.

[@lampad4549]

Why is the answer wrong?

[@charleswr8359]

9^x-6^x-4^x = 3^(2x)-3^x*2^x - 2^(2x) = 0 => divide by 2^(2x) (3/2)^(2x) - (3/2)^x - 1 = 0. Let y=(3/2)^x y^2 - y - 1 = 0.

Amazing

[@learncommunolizer]

Thanks 🙏❤️🙏

[@oswin627]

That's a perfect explanation.

[@learncommunolizer]

Thanks! Glad it was helpful!

[@취미생활-x2h]

what a beautiful math.

[@RixtronixLAB]

Nice video, thanks :)

[@learncommunolizer]

Thank you too!

[@poseidon.M]

Perfect🎉

[@owtwiepirate5238]

9^x-6^x=4^x ...(1) Divide (1) by 4^x [(3/2)^x]^2-(3/2)^x=1 ...(2) [(3/2)^x]^2-(3/2)^x-1=0 ...(3) (3/2)^x-1/2-(5)^(1/2)/2=0 ...(4) (3/2)^x-1/2+(5)^(1/2)/2=0 ...(5) But (5) is not satisfy (4): x= log[1/2+5^(1/2)/2] Note: log means log base 1.5

[@Granth_Sangrah]

Nice make rules are clear 👌

[@learncommunolizer]

Thanks 🙂

[@georgesadler7830]

Thank you for a well-organized math video.

[@learncommunolizer]

You're very welcome!

[@jjh4900]

1. Divide both sides by 6^x 2. (3/2)^x = X (X>0) 3. solve X^2-X-1=0 (X>0)

[@heraldomaciel9853]

Very Nice! I got too!

[@learncommunolizer]

Cool, thanks!

[@fredgandolfi2356]

Fun. Reminds me how much I enjoyed math in school and somewhat sadly how much I forgot since then.

[@yujunmomma2484]

I am not even interested in this math but i kept watching…

[@terratorment2940]

That's wild. My recollection of logarithms is a bit rusty. I probably never would have thought of those first few steps either

[@abeonthehill166]

Your explanation was so detailed and eloquent , i actually followed each step. Thanks for sharing……..

[@learncommunolizer]

Glad it was helpful! Your welcome!

[@erloos6791]

what is the pen are u using in this video sir can u pls reply me with link to buy this?

[@robertkapic6505]

Few steps ( log base 3/2) were unnecessary. But it was fun and reminded me of when I went to school.

[@rupertkeller2074]

Good work

[@learncommunolizer]

Thanks🙏

[@ОльгаСемина-з9л]

Мне 61... Но до первой половины все ясно... дальше темный лес... Во истину математика гимнастика 🤸‍♂️ для ума..

[@akibdekpa7069]

Nice ❤️❤️

[@learncommunolizer]

Thanks 🤗🥳

[@skyer.2299]

算那步一元二次方程为什么不直接配方？公式法岂不是算起来很麻烦？？

[@peelna2169]

I think if the step like this so long, the " trial and error method " is more suitable

[@stonejerome8291]

nice

[@learncommunolizer]

Thanks

[@ActMedInfo]

I dont undestand how log (1+√5) turned into log 3*236....

[@dashcam3683]

Thank you, these things almost scattered my brain in my high school days, just found out it was just fun and simple.

[@learncommunolizer]

You're very welcome!

[@rodicacojoc3636]

Explici lucrui simple cum sunt operaiile cu puteri sau acolo unde se aplica formula produsului de sumă a doi termen prin difența acelorași doi temeni , dar în final treci repede . De pildă ultimul logartm în ce bază era? Pobabil în baza 10, adică era logartm zecimal (pt care se pot lua valori din tabele sau de pe calculator) .După câte știu, logaritmul zecimal se noteaza cu lg.

[@DianaDawnDestiny]

My dad made me do this as practice when starting 8th grade.

[@ankitgupta-sn2xk]

how did you calculate the complex division and multiplication.. for example 9^1.486

[@shadow_1213]

With some abstract algebra applied to exponential equation: x = log((√5 - 1)/2)/log(2/3)

[@barsuk956]

Good 👍

[@learncommunolizer]

Thank you! Cheers!

[@mathpuzzles6352]

Good video, thanks

[@learncommunolizer]

Glad it helped

[@Hehehaw-tv4rs]

Really skillful, hope I will gain skill like you.

[@Kusy777ify]

Any problem solving way without using calculator ?

[@MrKockabilly]

Nice one. But please use "is equal to" not "is equals to". Sorry, I just noticed a lot (and I mean a lot) of people explaining calculations make that mistake

[@lorenzos.4442]

Very nice

[@learncommunolizer]

Thanks

[@florianvincent3932]

J’avais trouvé ! Yes ! Pensez à ramener à polynôme du second degré et ln est apparu la meilleure approche Le tout est de trouver le bon X! Hahaha