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

Question

I have a follow-up question to my last one. I need to solve for b for the following:

$$d=\frac{s(\ln(o))^b}{s(\ln(g))^b}$$

Once again, this is beyond the level of several online algebra calculators.

Answer 1

$$d=\frac{s(\ln(o))^b}{s(\ln(g))^b}$$ $$d= \left (\frac{\ln(o)}{\ln(g)} \right )^b $$ Taking logs of both sides ... $$ ln(d)=b \ln \left (\frac{\ln(o)}{\ln(g)} \right ) $$

$$ b = \frac{ \ln(d)} { \ln(\ln(o))-\ln(\ln(g))} $$

Answer 2

I interpreted $s(...)$ as being a function of one argument, as opposed to @WW1's answer where he interpreted it as a number multiplying both the numerator and the denominator of the fraction that could therefore be canceled. If my interpretation is correct, then

$$ d = \left(\frac{s(\ln(o))}{s(\ln(g))}\right)^b$$

Taking logs, we then get

$$ b = \frac{\ln(d)}{\ln(s(\ln(o)) - \ln(s(\ln(g))} $$

which cannot be simplified further.