How do I solve $\log(x^k) = x$ algebraically?

Question

I'm am trying to solve $\log(x^k) = x$ algebraically. But I am at a complete loss. Is this even possible with basic algebra?

Answer 1

The solution with the Lambert W function. By definition, $v=ue^u \Longleftrightarrow W(v) = u$. So formally: $$ \log(x^k) = x\\ x^k = e^x\\ x = e^{x/k}\\ 1 = xe^{-x/k}\\ \frac{-1}{k} = \frac{-x}{k}e^{-x/k}\\ W\left(\frac{-1}{k}\right) = \frac{-x}{k}\\ -kW\left(\frac{-1}{k}\right)=x $$

I said "formally" up there. I did not take into account the possibility of different $k$th roots, or different branches of the Lambert W.

added
Here are graphs of $\log(x^5)$ and $x$.

And our solution with the two real branches of the W function: $$ -5 W_0\left(\frac{-1}{5}\right) \approx 1.296\\ -5 W_{-1}\left(\frac{-1}{5}\right) \approx 12.71 $$ Complex non-real solutions are obtained using the other branches of W.

Answer 2

take the LambertW function $$x=-k{\rm W} \left(-{k}^{-1}\right)$$

Answer 3

It's not possible with what most people call basic algebra, but if you allow the use of the Lambert W function it is possible.