1. type of exponential equation

To solve $2^x = x^5$ could you use Lambert W function?

Thanks

2. It can be solved via derivatives and intersections of curves. But how would you solve it in a purely algebraic manner using the W Lambert function?

3. Originally Posted by tukeywilliams
It can be solved via derivatives and intersections of curves. But how would you solve it in a purely algebraic manner using the W Lambert function?

In principle yes it can, the equation can be transformormrd to:

$
\frac{-x \ln(2)}{5}e^{\frac{-x \ln(2)}{5}}=\frac{- \ln(2)}{5}
$

So formaly we have:

$
\frac{-x \ln(2)}{5} = W\left(\frac{- \ln(2)}{5}\right) \approx W(-0.1386)
$

Now the problem is that there a multiple (real) branches of the Lambert W in
the region around $-0.1386$, and you are going to have to
evaluate two of them to get all the solution to this. My LambertW calculator
gives $x\approx 1.17724$.

However just plot the functions and you will see that there is another
solution near $x=22$ Numerical methods will find both solutions
once you know where they are.

RonL

4. Here is a graph if you'd like to see...

5. Originally Posted by CaptainBlack

...there is another solution near $x=22$...

RonL
I really couldn't see that one when looking at the graph...

6. Originally Posted by janvdl
I really couldn't see that one when looking at the graph...
A binary search gives me about x = 22.440006

-Dan