Solving for a variable in an algebra equation with logs and powers

Question

I'm working on a software project and I need to solve for b in the following equation:

$$e^{\ln(gx)ab}-e^{\ln(x)ab}=c$$

I've tried a couple of online algebra calculators, but they cannot resolve this.

Answer 1

$c=e^{\ln(gx)ab}-e^{\ln(x)ab}=gx^{ab}-x^{ab}=x^{ab}(g-1)$, hence

$x^{ab}=\frac{c}{g-1}$, thus

$ab \ln (x)= \ln (c)- \ln (g-1)$.

Can you proceed ?