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.
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.
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.
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.
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)
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!
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. .
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)
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.
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
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.
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.
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
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.
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)
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.
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.
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.
@@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 😶
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.
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.
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.
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.
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.
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
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.
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
9^x - 6^x = 4^x Let’s start by dividing both sides by 9^x: 1 - (6/9)^x = (4/9)^x Let y = (4/9)^x: 1 - y = (2/3)y y = 3/5 (4/9)^x = 3/5 Take the logarithm of both sides: log(4/9)^x = log(3/5) xlog(4/9) = log(3/5) x = log(3/5) / log(4/9) Therefore, x ≈ 0.847.
Actually, there are 2 more solutions: one of them follows from negative solution of that quadratic equation (it will contain imaginary part), and the second one is minus infinity.
@@coolcodingcat The constraints are implicit, but yeah I can agree that we can go all technical about this, but the general rule is that default/implied constraints applies when they are not specifically stated. Put up a polynomial with integers without constraints and parameters, it would range from unwise to incorrect to involve complex numbers or other fields of mathematics not specifically invoked by explicitly stated constraints.
@@gandalf8216 the constraints are not implied, you only assumed the constraints. And it is neither unwise nor incorrect to include complex numbers when the domain is not specified. The only reasons the video does not include them was either for simplicity or because he was unaware of them. The domain may be assumed to be limited to real numbers when dealing with lower level math or to not confuse people who maybe have not learned about imaginary numbers or complex numbers, but in a pure mathematical sense, imaginary numbers and complex numbers are just as valid as real numbers.
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
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.
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
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
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 ;)
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
Listening this solution, I forgot the math I knew, but I learned Indian pronunciation. Thank you so much!!!
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!
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
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!
Very instructive task
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)
great and simple teaching. thanks.
You are welcome! 🙏❤️🙏
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
very interesting approach, many thank for this problem
Thanks and Welcome 🙏❤️🙏
Thanks 😊... This is Abracadabra 😊🌟💌👈9™-6™=3™...TM=1... This is True.
Thanks 😊...
You're welcome 😊
Exponentials...quadratics...logs....Such a simple looking question and yet amazing
Really nice 👌
Good
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.
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
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.
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.
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
Excellent question, and good explanation, thanks.
You are welcome!🙏❤️
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.
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)
Very nice congratulations👏👏👏
🤝👍👍
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.
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 🙏❤️
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?
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
Very nice explanation
Thanks and welcome
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 😶
Good explanation
Thanks and welcome
Amazing
Thanks 🙏❤️🙏
Perfect🎉
Math skills refreshed, thanx...
Problems like this are essentially formulaic. So ln(golden ratio)/ln(3/2).
my man just solved creation of universe
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
1. Divide both sides by 6^x
2. (3/2)^x = X (X>0)
3. solve X^2-X-1=0 (X>0)
nice
Thanks
Very good explanation going through all the steps. Thanks! 👍
You're 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.
Watching that without sound. Looked like a satisfying video.
Glad you enjoyed it!
Brilliant. . . . . . .
Excellent. . . . . . . .
Nonsense ॥
Thank make more videos. Good luck 👍❤❤
Thank you, I will try my best 👍❤️❤️❤️ @Joysaha
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
Very nice explanation.
Because of this video I have revised 3 chapters, quadratic equations, logarithms, exponents
Excellent.All the best!
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.
i like how most kept on watching without having a clue of what is going on, like me.
Great explanation! Towards the end of your video, I think I heard a cat purring. 🤣
😁 You are right !
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 ✌️
Great job.
I would consider the problem solved at 8:41
Your explanation was so detailed and eloquent , i actually followed each step.
Thanks for sharing……..
Glad it was helpful!
Your welcome!
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.
Beautiful problem, but not tough enough for Olympiad
9z
Zz
z😊si 😊z
00p😅
Nice que. Thanks bro
Thanks too! Your Welcome!
What a long and beutiful journey
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
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
Nice ❤️❤️
Thanks 🤗🥳
Very good content.
Much appreciated &
Glad you liked it!
Very impressive. Nice work.
Thank you! Cheers!
I do admire and feel the sincerity of your content. More success and support to you. Cheers. 💕
Thank you so much 🤗 @wowcheers7488
Good work
Thanks🙏
Very nice
Thanks
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.
what a beautiful math.
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 like the way you present it, Bravo!!
Asrrf
Hjp😅
Jgj
Good 👍
Thank you! Cheers!
Nice
Thanks
I think if the step like this so long, the " trial and error method " is more suitable
Мне 61... Но до первой половины все ясно... дальше темный лес... Во истину математика гимнастика 🤸♂️ для ума..
👍🏼
It was refreshing...love your accent
Numerical calculation is... not required. It makes problem more complex.
Vrlo konplikovano,veoma nejasno objašnjenj
Nice make rules are clear 👌
Thanks 🙂
My dad made me do this as practice when starting 8th grade.
Thank you for a well-organized math video.
You're very welcome!
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.
well done! good jo!
Thank you! 😃
Good video, thanks
Glad it helped
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)
That's a perfect explanation.
Thanks! Glad it was helpful!
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.
Fun. Reminds me how much I enjoyed math in school and somewhat sadly how much I forgot since then.
Any problem solving way without using calculator ?
9^x - 6^x = 4^x
Let’s start by dividing both sides by 9^x:
1 - (6/9)^x = (4/9)^x
Let y = (4/9)^x:
1 - y = (2/3)y
y = 3/5
(4/9)^x = 3/5
Take the logarithm of both sides:
log(4/9)^x = log(3/5)
xlog(4/9) = log(3/5)
x = log(3/5) / log(4/9)
Therefore, x ≈ 0.847.
That's wild. My recollection of logarithms is a bit rusty. I probably never would have thought of those first few steps either
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？？
Actually, there are 2 more solutions: one of them follows from negative solution of that quadratic equation (it will contain imaginary part), and the second one is minus infinity.
Neither of those are real numbers, and so are outside the domain of the equation.
@@gandalf8216 No, you are wrong. The domain of the equation is not constrained, because no constraint was stated for the value of x.
@@coolcodingcat The constraints are implicit, but yeah I can agree that we can go all technical about this, but the general rule is that default/implied constraints applies when they are not specifically stated. Put up a polynomial with integers without constraints and parameters, it would range from unwise to incorrect to involve complex numbers or other fields of mathematics not specifically invoked by explicitly stated constraints.
@@gandalf8216 the constraints are not implied, you only assumed the constraints. And it is neither unwise nor incorrect to include complex numbers when the domain is not specified. The only reasons the video does not include them was either for simplicity or because he was unaware of them. The domain may be assumed to be limited to real numbers when dealing with lower level math or to not confuse people who maybe have not learned about imaginary numbers or complex numbers, but in a pure mathematical sense, imaginary numbers and complex numbers are just as valid as real numbers.
"Minus infinity" isn't a solution.