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.
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.
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.
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.
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.
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.
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.
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.
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!
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)
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. .
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)
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.
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.
@@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
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)
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.
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
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!
@@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.
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
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
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.
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.
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.
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.
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.
@@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
@@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 😶
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
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.
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.
@@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
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.
making a program to solve this is a easy, even with basic math knowlege you can take a few assumptions 1: X must be small or the 9^x will dominate an the left side will be bigger 2:if X is too small then the left side is smaller so need to increase X, if left side is bigger then decrease X 3:X is probably a decimal or it could be just brute forced easily So with those knowlege just make a binary search for the X and the end range is the first X where left side is bigger, just make it search also for decimal numbers. For me it resulted in X = 1.1868143902809818, but here float inaccuracy plays a big role, but I was too lazy to make it more accurate.
Hello, could you please explain the part starting at 2:57 where you introduced a,b and c as well as the new square root equation? I’m a little confused as to the values if a,b and c as well as the root equation formula. Thank you!
ax^2 + bx + c = 0. The quadratic equation solution can be obtained my using a method known as completing the square and get an intermediate equation along the lines of: (x-u)^2 + v = 0 And then solving for x.
What Olympiad would this be from? I did nationally competitive math and Olympiad problems usually cause some consternation, but this I chugged out with ease. I disagree with the characterization that this is an Olympiad-style question. This is on par with a high school algebra 2 bonus test question
I don't think it's really necessary to go all out with the calculations of the logs. The answer of x = log(something) / log(something) is perfectly acceptable, since it's only calculations made with the help of a calculator that are required from that step on.
Hello. You should not switch to decimal approximation when you have the solution. Mathematics doesn't care. The true solution is X = log_3/2 ((1+sqrt(5))/2)
I don't bother with long division anymore since they invented calculators. And I don't bother with algebra anymore since they invented Wolfram Alpha. You can just plug the answer there and get the answer in about two seconds, including some of the work seen in this video.
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.
I need to make a log
@@margarita8442 what is log i heard that a lot in some kind of movies. "Log logg log log mmh "
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
@@serkan5951 what is log? Baby don't hurt me, don't hurt me, no more
2
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.
Same thoughts
The golden ratio is also very important in other fields, such as architecture and music.
φ is the letter you are looking for
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
Please explain more
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.
You approximate 3times You have to approximate o će finaly
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.
math olympiad has catagories remember?
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.
Yes I did similar but I completed the square instead of using quadratic formula.
Wrong, you cant do a second grade equation if a is not by multiplied by X²
@@ashamenai1659 The quadratic formula is an application of completing the square ;)
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.
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
Listening this solution, I forgot the math I knew, but I learned Indian pronunciation. Thank you so much!!!
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.
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.
Try to divide by 6^x, or by 9^x. They're all the same
@davidzhang2319 you will require calculus to do that.
As for this problem you can divide by any. I divided by 6
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!
Glad you enjoyed it!
I was doing fine until the log showed up
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)
very interesting approach, many thank for this problem
Thanks and Welcome 🙏❤️🙏
Exponentials...quadratics...logs....Such a simple looking question and yet amazing
Way easier answer………
A= 2021, B= 0, C= 2020. Plug that in for both equations and it works for both.
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.
.
Using code is too overpowered
@@thetimeiable coding rocks!
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)
Exactly
watch this nice math problem : ruclips.net/video/3qbtT50Ag0k/видео.html
indeed!
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.
great and simple teaching. thanks.
You are welcome! 🙏❤️🙏
Excellent question, and good explanation, thanks.
You are welcome!🙏❤️
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.
This magic man really did do something magical here! So many mathematical atrocities! Elegant!
@@arnavkmr3895 It's was a joke ;)
@@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
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)
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.
Thanks 🙏❤️
Thanks 😊... This is Abracadabra 😊🌟💌👈9™-6™=3™...TM=1... This is True.
Thanks 😊...
You're welcome 😊
Thank make more videos. Good luck 👍❤❤
Thank you, I will try my best 👍❤️❤️❤️ @Joysaha
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
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!
@@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.
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
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
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.
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
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.
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.
12:59
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.
Very nice congratulations👏👏👏
🤝👍👍
Very good explanation going through all the steps. Thanks! 👍
You're welcome!👍👍👍
Very impressive. Nice work.
Thank you! Cheers!
Very instructive task
Very nice explanation.
Because of this video I have revised 3 chapters, quadratic equations, logarithms, exponents
Excellent.All the best!
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.
Yeah. I’m surprised I Rem the log rules even though I learnt them 23 years ago.
@@lordndrew Wow, 23 years and its still all there too?
Nine years in my case
Bringing back my school memories, we use to do these in 6 to 8 th grades
Fekuu 😆😆
Kya fek raha hain bhai,jee mains se tougher sawaal hai yeh sab
@@greninja7619 nhi hai shayd tumne jee mains ke questions nhi dekhe
@@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
@@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 😶
Great explanation! Towards the end of your video, I think I heard a cat purring. 🤣
😁 You are right !
I like the way you present it, Bravo!!
Asrrf
Hjp😅
Jgj
Really nice 👌
my man just solved creation of universe
i like how most kept on watching without having a clue of what is going on, like me.
I do admire and feel the sincerity of your content. More success and support to you. Cheers. 💕
Thank you so much 🤗 @wowcheers7488
1. Divide both sides by 6^x
2. (3/2)^x = X (X>0)
3. solve X^2-X-1=0 (X>0)
Cant beleive every student learnt how to solve this problem at highschool as a easy question in my country.
Great job.
I would consider the problem solved at 8:41
Great explanation! Very interesting problem
Glad it was helpful!
I think you can final in (log(1+sqrt(5)/2)/(log(3/2)), this is a answer it most accurate
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
11:45 how did you count the values ? shortened this detail but how ? 👎
It was refreshing...love your accent
Watching that without sound. Looked like a satisfying video.
Glad you enjoyed it!
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.
Thanks and Welcome ✌️
Very nice explanation
Thanks and welcome
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.
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.
watch this nice logarithmic problem : ruclips.net/video/3qbtT50Ag0k/видео.html
Beautiful content bro,but this ain't Olympiad material✌️♥️
Yes it is
@@learncommunolizer q@
QAa@@learncommunolizer aaa
Aa11
@@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
Beautiful problem, but not tough enough for Olympiad
9z
Zz
z😊si 😊z
00p😅
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.
Why is the answer wrong?
Good explanation
Thanks and welcome
Math skills refreshed, thanx...
Thank you for a well-organized math video.
You're very welcome!
making a program to solve this is a easy, even with basic math knowlege you can take a few assumptions
1: X must be small or the 9^x will dominate an the left side will be bigger
2:if X is too small then the left side is smaller so need to increase X, if left side is bigger then decrease X
3:X is probably a decimal or it could be just brute forced easily
So with those knowlege just make a binary search for the X and the end range is the first X where left side is bigger, just make it search also for decimal numbers.
For me it resulted in X = 1.1868143902809818, but here float inaccuracy plays a big role, but I was too lazy to make it more accurate.
what is the pen are u using in this video sir can u pls reply me with link to buy this?
Numerical calculation is... not required. It makes problem more complex.
how did you calculate the complex division and multiplication.. for example 9^1.486
Nice que. Thanks bro
Thanks too! Your Welcome!
Your explanation was so detailed and eloquent , i actually followed each step.
Thanks for sharing……..
Glad it was helpful!
Your welcome!
Hello, could you please explain the part starting at 2:57 where you introduced a,b and c as well as the new square root equation? I’m a little confused as to the values if a,b and c as well as the root equation formula. Thank you!
ax²+bx+2 standard form, a,b,c also known as constant or coefficient
That formula with a, b and c is called the quadratic equation.
ax^2 + bx + c = 0. The quadratic equation solution can be obtained my using a method known as completing the square and get an intermediate equation along the lines of:
(x-u)^2 + v = 0
And then solving for x.
Very good content.
Much appreciated &
Glad you liked it!
First of all there are mcq in olympiads ....so make a easy way
.... for eg.... make it 4/3^x... and 3^x-2^x.... from here 1
I am not even interested in this math but i kept watching…
With some abstract algebra applied to exponential equation:
x = log((√5 - 1)/2)/log(2/3)
Can we not use log directly from the beginning ?
Мне 61... Но до первой половины все ясно... дальше темный лес... Во истину математика гимнастика 🤸♂️ для ума..
Any problem solving way without using calculator ?
Problems like this are essentially formulaic. So ln(golden ratio)/ln(3/2).
What Olympiad would this be from? I did nationally competitive math and Olympiad problems usually cause some consternation, but this I chugged out with ease. I disagree with the characterization that this is an Olympiad-style question. This is on par with a high school algebra 2 bonus test question
Well done. Rounding is so unsatisfying after all that. Would have been nice to get a whole number.
What does the i mean in quadratic equations?
That's wild. My recollection of logarithms is a bit rusty. I probably never would have thought of those first few steps either
I just had a Math exam last week, I can't believe I'm watching this video for entertainment..
I don't think it's really necessary to go all out with the calculations of the logs. The answer of x = log(something) / log(something) is perfectly acceptable, since it's only calculations made with the help of a calculator that are required from that step on.
Few steps ( log base 3/2) were unnecessary. But it was fun and reminded me of when I went to school.
Hello. You should not switch to decimal approximation when you have the solution. Mathematics doesn't care.
The true solution is X = log_3/2 ((1+sqrt(5))/2)
I'd state it more strongly: mathematics doesn't want an approximation unless you see a specific request for an approximation.
U may do apply " squaring both the side"in start.
Do you know the intrinsic value of 1 plus 1?
I still haven't figured out solving log problems. I'm 66 but desiring to learn.
Nice ❤️❤️
Thanks 🤗🥳
Perfect🎉
If you take logs at start you can easily get to x=1/log(1.5) in your head.
Nice video, thanks :)
Thank you too!
❤❤❤
❤️❤️❤️@ThayNguyenManhHung
I don't bother with long division anymore since they invented calculators. And I don't bother with algebra anymore since they invented Wolfram Alpha. You can just plug the answer there and get the answer in about two seconds, including some of the work seen in this video.
Good work
Thanks🙏
Brilliant. . . . . . .
Excellent. . . . . . . .
Nonsense ॥
My dad made me do this as practice when starting 8th grade.
I just wonder why it is olympiad math problem. Anyway your solution would be helpful for high school students. Thanks.
Happy to help
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？？