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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.

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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 $$z_\pm=z_*\pm \sqrt{-2\frac {f(z_*)}{f''(z_*)}}$$

Then use Newton method to polish the root(s) if any does exist in the real domain.

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