# 2^x=x^5 anyway to solve this?

EternityMech
anyway to solve this? if so what level of math is needed?

## Answers and Replies

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EternityMech

what if you logged both sides using log with 2 power base?

you get x = 5 lnx

is that easier to solve?

anyway to solve this? if so what level of math is needed?

As micromass said, you can't really solve that type of equation algebraically. In some simple case you might just "spot" a solution by inspection (eg 2^x = x^2; x=2 or x=4).

You can solve the equation numerically or manipulate the Lambert-W function into a solution.

The Lambert-W function W(x) is defined as the solution (w) to $we^w = x$.

BTW. How many real solutions are you expecting to get for that equation?

what if you logged both sides using log with 2 power base?

you get x = 5 lnx

is that easier to solve?

No, there's still no algebraic solution.

You're best to write it as $e^{ax} = x^5$ and then take the fifth root of each side. (a = sqrt(2) btw).

$x e^{-ax/5} = 1$

From here you can fairly easily manipulate it into Lambert's equation.

EternityMech

BTW. How many real solutions are you expecting to get for that equation?

all of them...?

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all of them...?

Can you deduce how many there are?

EternityMech

Can you deduce how many there are?

2 if you use calculus it seems atleast according to wolfram. been looking into the lambert w function seems waay out of my league for now. seems there are no simple answer with this. but it sure looks simple.

Reptillian

Solve it graphically! A picture is worth a thousand words.

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2 if you use calculus it seems atleast according to wolfram. been looking into the lambert w function seems waay out of my league for now. seems there are no simple answer with this. but it sure looks simple.

$x=\cos(x)$ might also look simple, but it too cannot be solved algebraically. They're called transcendental equations.

2 if you use calculus it seems atleast according to wolfram. been looking into the lambert w function seems waay out of my league for now. seems there are no simple answer with this. but it sure looks simple.
Yes that is correct, there are two real solutions. For some fairly small positive value of "x", x^5 will exceed 2^x, but it is important to understand that for large "x" that 2^x will eventually exceed x^5 (or any power of x for that matter).

BTW. From my previous post $x e^{-ax/5} = 1$, so $(-ax/5) e^{-ax/5} = -a/5$. Can you see that this is exactly in the form of Lambert's equation. Of course this is only of use to you if you have software or tables or whatever that solves Lambert's equation.

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Rearrange the equation to $x = (2^x)^{0.2}$ and start with an initial "guess" of say x=2. You'll find that repeated iterations of this equation converge to the first solution fairly quickly. This is called "fixed point iteration" btw.
For the second solution you can rearrange it into $x = \log(x^5)/\log(2)$, with some larger initial guess for "x" this should converge fairly quickly to the second solution.