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

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.

• Is this equivalent to $e^{\alpha b}-e^{\beta b}=c$ ? If it is, except in very few cases where this can reduce to a polynomial, consider numerical methods. – Claude Leibovici Aug 10 '18 at 10:08

$c=e^{\ln(gx)ab}-e^{\ln(x)ab}=gx^{ab}-x^{ab}=x^{ab}(g-1)$, hence
$x^{ab}=\frac{c}{g-1}$, thus
$ab \ln (x)= \ln (c)- \ln (g-1)$.
• I can solve for b with that. Let me test it... – IamIC Aug 10 '18 at 10:13