Given: $a_i>0$, $b_1>0$, $b_2<0$
Solve for: $a_1\ln(1+b_1x)+a_2\ln(1+b_2x)=0$
My try: The equation is equivalent to $(1+b_1x)^{a_1}=(1+b_2x)^{-a_2}$,
Which seems to be solved with only non-elementary functions...
Is it possible to prove that no elementary function can solve this and then find a non-elementary one to solve this?
You want to find the zero's of function $$f(x)=a_1\log(1+b_1x)+a_2\log(1+b_2x)$$ This is an highly transcendental equation and there is no hope to find an explicit solution if $a_1 \neq a_2$; this means that you will need some iterative numerical methods and then a reasonable starting guess.
For the first derivatives, we have $$f'(x)=\frac {a_1b_1 }{1+b_1x }+\frac {a_2b_2 }{1+b_2x }$$ $$f''(x)=-\frac {a_1b_1^2 }{(1+b_1x)^2 }-\frac {a_2b_2^2 }{(1+b_2x)^2 }\quad <0 \quad \forall x$$
The first derivative cancels at $$x_*=-\frac {a_1b_1 +a_2b_2 } {(a_1+a_2)b_1b_ 2}$$ which may exist (the requirement is $x < -\frac 1{b_2}$ because of the logarithm). If it does and $f(x_*)>0$, there will two roots since, by the second derivative test, $x_*$ corresponds to a maximum. But $x=0$ being a solution, we need the other one in the reduced range $$-\frac 1 {b_1} < x < x_*$$
Make a series expansion of $g(x)$ around $x_*$ $$f(x)=f(x_*)+\frac{1}{2} f''(x_*)(x-x_*)^2 +O\left((x-x_*)^3\right)$$ and solve it to get, as an approximation, $$x_0=x_*- \sqrt{-2 \frac{f(x_*)}{f''(x_*)}}$$
Let us try with $a_1=1$, $b_1=3$, $a_2=5$, $b_2=-7$. These conditions give $$x_*=-\frac{16}{63} \qquad \text{and} \qquad f(x_*)=5 \log \left(\frac{25}{9}\right)-\log \left(\frac{21}{5}\right) >0$$ but is gives $$x_0=-\frac{16}{63}-\frac{5}{63} \sqrt{\frac{5}{3} \left(5 \log \left(\frac{25}{9}\right)-\log \left(\frac{21}{5}\right)\right)}\approx -0.450338$$ which is outside the range $\left(-\frac 13,-\frac{16}{63} \right)$; so, this does not work but we reduced significantly the range to be explored since now we know that the solution is such that $$-\frac 13 <x<-\frac{16}{63}$$
So, at least for the time being, what I would suggest is to divide the new interval (say $100$ equally spaced points) and pick as a starting guess the last value of $x$ for which $f(x) <0$ (application of Darboux theorem).
For the worked example, we shall have to compute the $f(x_i)$ $$x_i=-\frac 13+\frac 1 {100}\left(-\frac{16}{63}+\frac 13\right) i=\frac{i}{1260}-\frac{1}{3}$$ and $x_1=-\frac{419}{1260}$ becomes our starting point. Now, using (as the simplest) Newton method, the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & -0.3325396825 \\ 1 & -0.3325166882 \\ 2 & -0.3325163490 \end{array} \right)$$
What we could try is one iteration of the quintic order of Newton-like method starting with $x_0=x_*$; this order is required being the first which makes the function value appearing in denominator.
It will write $$x_1=x_*-\frac{4 \,f(x_*)\, f'''(x_*)}{6 f''(x_*)^2-f(x_*) \,f''''(x_*)}$$
Applied to the worked problem, it gives $$x_1=\frac{32 \left(45+220 \log \left(\frac{25}{9}\right)-44 \log \left(\frac{21}{5}\right)\right)}{567 \left(10+35 \log \left(\frac{25}{9}\right)-7 \log \left(\frac{21}{5}\right)\right)}\approx -0.326529$$