- #1

- 36

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a

^{x}= x

^{b}

It kinda tripped me out at first.

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- Thread starter JungleJesus
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- #1

- 36

- 0

a

It kinda tripped me out at first.

- #2

nicksauce

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There is no way in general to solve this kind of equation in terms of elementary functions.

- #3

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You can try manipulating it to fit the lambert W function I think.

- #4

HallsofIvy

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Yes. Taking the "b"th root of both sides, [itex](a^{1/b})^x= x[/itex] so that [itex]1= xe^{-(1/b)(ln a)x}[/itex]. Multiply on both sides by -(1/b)ln a: [itex]-(1/b)ln a[/itex][itex]= (-(1/b)(ln a)x)e^{-(1/b)(ln a)x}[/itex].You can try manipulating it to fit the lambert W function I think.

Letting y= -(1/b)(ln a)x, that becomes [itex]ye^y= -(1/b)ln a[/itex]. Using the Lambert W function (which is defined as the inverse function to f(x)= xe

Then [itex]x= -\frac{b}{ln a}W(-(1/b)ln a)[/itex]

- #5

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Letting y= -(1/b)(ln a)x, that becomes [itex]ye^y= -(1/b)ln a[/itex]. Using the Lambert W function (which is defined as the inverse function to f(x)= xe^{x}) we have y= W(-(1/b)ln a).

Then [itex]x= -\frac{b}{ln a}W(-(1/b)ln a)[/itex]

Is there any way to simplify this expression, or is this the bare bones? How would one procede to calculate this value?

- #6

CRGreathouse

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Is there any way to simplify this expression, or is this the bare bones? How would one procede to calculate this value?

There are numerical methods, like Newton's method or the secant method, that can give numerical values for the W function -- or even your original function directly. (If you just want the W function, there are optimizations that can be used to make it go faster.)

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