# Solve $\left(v^{\log _b\left(a-m\right)}+m\right)^z+m^z=a-m$

While constructing some exponential curve, I have been trying to solve the following:

$$\left(v^{\log_{b}\left(a-m\right)}+m\right)^{z}+m-m^{z}=a$$

for $$z$$. I have tried using WolframAlpha, but it was unable to get any answer for $$z$$ within the standard computation limit, while Symbolab couldn't arrive at any answer. I'm not sure how to solve this by hand. I would initially try to reduce it to the form

$$\left(v^{\log _b\left(a-m\right)}+m\right)^z-m^z=a-m$$

but now, I'm not sure how to reduce this any further to get an answer for $$z$$.

In context, I can assume that $$v$$, $$b$$, $$a$$ and $$m$$ are positive real numbers, and that $$a > m$$.

I couldn't seem to be able to apply any log rules (since the bases would be different; e.g. $$\log(x)+\log(y)=\log(xy)$$ only holds when both $$\log(x)$$ and $$\log(y)$$ were to have the same base, which in this case they don't), so I don't know how to continue.

I suppose (and more) that you will need some numerical method; however, there a few things that xe could do considering that we look for the zero's of function $$f(z)=c ^z-m ^z-d \quad \text{where} \quad c=\left(v^{\log _b\left(a-m\right)}+m\right)\quad \text{and} \quad d=(a-m)$$ We have $$f'(z)=c^z \log (c)-m^z \log (m)\quad \text{and} \quad f''(z)=c^z \log ^2(c)-m^z \log ^2(m)$$ The first derivative cancels at $$z_*=-\frac{\log \left(\frac{\log (c)}{\log (m)}\right)}{\log (c)-\log (m)}$$ At this point, we need to assume that $$f(z_*) <0 \quad \text{if} \quad f''(z_*) >0$$ or that $$f(z_*) >0 \quad \text{if} \quad f''(z_*) <0$$
If this is the case, for an approximation, you could build a Taylor expansion around $$z_*$$ and get as possible estimates $$z_\pm=z_*\pm \sqrt{-2\frac {f(z_*)}{f''(z_*)}}$$