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).
On
$$a^xd_2^x+b^\frac{x}{2}d_1^x-d_1^xd_2^x=0$$
In the general case, the summands on the left-hand side are algebraically independent. This equation is therefore a polynomial equation of more than one algebraically independent monomials ($a^xd_2^x,b^\frac{x}{2}d_1^x,d_1^xd_2^x$) and with no univariate factor. We therefore don't know how to solve the equation by rearranging for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.
$$a^xd_2^x+(\sqrt{b})^xd_1^x-d_1^xd_2^x=0$$ $ad_2\to u$, $\sqrt{b}d_1\to v$, $d_1d_2\to w$: $$u^x+v^x-w^x=0\tag{1}$$
For real $a,b,d_1,d_2,x$, we can transform this equation to a trinomial equation with real exponents:
$x\to\ln(t)$ ($t>0$): $$t^{\ln(u)}+t^{\ln(v)}-t^{\ln(w)}=0$$
We can transform this equation to the form of equation 8.1 of [Belkic 2019]:
$$t^{\ln(u)-\ln(w)}+t^{\ln(v)-\ln(w)}-1=0$$ $t\to z^\frac{1}{\ln(u)-\ln(w)}$: $$z+z^{\frac{\ln(v)-\ln(w)}{\ln(u)-\ln(w)}}-1=0$$
Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $\ _1\Psi_1$ can be obtained therefore.
$\ $
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