How can I solve an exponential equation of the following type? I have an equation of the form 
$$
\frac{a^x}{d_1^x} + \frac{b^{x/2}}{d_2^x} = 1,
$$
which I have already rewritten to
$$
a^xd_2^x+d_1^xb^{x/2}-d_1^xd_2^x = 0.
$$
However, I seem to be stuck here. Howe can I solve this equation for $x$? 
In case, that's important: $d_1$ and $d_2$ can be considered constant, while $a$ and $b$ can each be measured multiple times. (Having something like an (overdetermined) equation system in mind).
 A: Hoping that $b>0$, I think it would be simpler to rewrite $$\frac{a^x}{d_1^x} + \frac{b^{x/2}}{d_2^x}= \Big(\frac{a}{d_1}\Big)^x+\Big(\frac{\sqrt b}{d_2}\Big)^x$$ and so the equation to solve is looking like $$c_1^x+c_2^x=1$$ for which there is no analytical solution.
As I said in comments, only numerical methods could solve the problem. The simplest would be Newton which, starting from a reasonable guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For your problem (as I rewrote it) $$f(x)=c_1^x+c_2^x-1$$ $$f'(x)=c_1^x \log (c_1)+c_2^x \log (c_2)$$
For illustartion purposes, let us use $c_1=2,c_2=3,x_0=0$. The successive iterates will then be generated : $-0.558111$, $-0.765304$, $-0.787654$, $-0.787885$ which is the solution for six significant figures.
Added later
This function $$f(x)=c_1^x+c_2^x-1$$ is interesting to solve because its second derivative $$f''(x)=c_1^x \log ^2(c_1)+c_2^x \log ^2(c_2)$$ is always positive. So, at most, the equation has two roots. 
Moreover $f(0)=1$ and $f''(0) >0$ which implies that, starting iterations at $x_0=0$, we should always go to the root (if it exists) without any overshoot (as shown in the example above); so the appropriate starting point should be $x_0=-\frac{1}{\log ({c_1 c_2})}$. The first derivative cancels for $$x_*=-\frac{\log \left(-\frac{\log (c_1)}{\log (c_2)}\right)}{\log (c_1)-\log (c_2)}$$ and if $f(x_*) <0$ a second root exists
