How do you solve
$$\frac{6^x-1}{8^x-1}=\frac{3}{4}$$ so $x$ must not be $0$. Using some algebra I could simplify it into $4×6^x-3×8^x-1=0$.
I don't know what I should do after this. I put this exercise into Wolfram Alpha I get the approximate of $0.77$.
The ratio $f(x)=\frac{6^x-1}{8^x-1} $ varies almost linearly around $x=1$. So, a decent approximation can be obtain with
$$x= 1+\frac{ f(1)}{f’(1)} =1+\frac{\frac1{24}}{\ln3-\frac{13}7\ln2}= 0.779$$