$a*b = k*a^b$, solve for b?

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I was working on an open ended math puzzle and I ended up with this equation. Is there a way to solve for it analytically without graphing?

I got stuck on trying to figure out how to find $b$ if $b$ times a constant $c$ is equal to $a$ times itself $b-1$ times. Also, if it helps, $b$ mod $a$ is not guaranteed to not be 0, and b must be at least 1.

If $a*b = k*a^b$, find b.

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There is no way to solve this in elementary terms. However, we may use the Lambert W function to solve for $b$ as such:$$\begin{align*}ab&=ka^b\\ba^{-b} &= \frac{k}{a}\\-be^{-b\ln{a}} &= \frac{-k}{a}\\-b\ln{a}e^{-b\ln{a}} &= \frac{-k\ln{a}}{a}\\-b\ln{a}&=W\left(\frac{-k\ln{a}}{a}\right)\\b &= \frac{-W\left(\frac{-k\ln{a}}{a}\right)}{\ln{a}}\end{align*}$$