A Nice Math Olympiad Algebra Problem | How to solve!!

In todays video, I will be teaching you how to solve an interesting problem. Make sure to like, subscribe, and also comment any questions or video ideas you may have relating to math!
#maths #exponents #mathematics #mathstricks #mathchallenge #algebra #equation #problem #challenge #Olympiad #MathOlympiad

[@SttmpJohnson]

There is also a solution where x is approximately equal to 1.0224.

[@payoo_2674]

Also where x is approximately equal to -0.97902.

[@viavip1940]

when the comments solve better than the video

[@ernestosaez8058]

You are not solving the equation, just finding one solution which is, let’s say, intuitive. But this method will not always work, so it’s not a valid way to solve the equation.
There are, in fact, two other solutions.

[@payoo_2674]

Use the Lambert W function W(■*e^■) = ■
2^x = x^32
ln(2^x) = ln(x^32)
x*ln(2) = 32*ln|x| ===> two cases
1st case: x > 0
x*ln(2) = 32*ln(x)
ln(x)*x^(-1) = ln(2)/32
ln(x)*e^ln(x^(-1)) = ln(2)/32
ln(x)*e^(-ln(x)) = ln(2)/32
-ln(x)*e^(-ln(x)) = -ln(2)/32
W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32)
-ln(x) = W(-ln(2)/32)
ln(x) = -W(-ln(2)/32)
x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
x1 = e^(-W_[0](-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
in WolframAlpha: e^(-productlog(0,-ln(2)/32))
x2 = e^(-W_[-1](-ln(2)/32)) = 265
in WolframAlpha: e^(-productlog(-1,-ln(2)/32))
2nd case: x < 0
x*ln(2) = 32*ln(-x)
ln(-x)*x^(-1) = ln(2)/32
-ln(-x)*x^(-1) = -ln(2)/32
ln(-x)*(-x)^(-1) = -ln(2)/32
ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32
ln(-x)*e^(-ln(-x)) = -ln(2)/32
-ln(-x)*e^(-ln(-x)) = ln(2)/32
W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32)
-ln(-x) = W(ln(2)/32)
ln(-x) = -W(ln(2)/32)
-x = e^(-W(ln(2)/32))
x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
x3 = -e^(-W_[0](ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...
in WolframAlpha: -e^(-productlog(0,ln(2)/32))