How to solve $\frac{6^x-1}{8^x-1}=\frac{3}{4}$

Question

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 $0.77$.

That $0.77704...$ is the result of numerical methods, which is why it says in WA "numerical solution". There is most likely no way to solve it algebraically. — Micah Windsor, May 06 '20 at 13:56

score 1 · Answer 1 · answered May 06 '20 at 19:50

1

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

$$x= 1+\frac{ f(1)}{f’(1)} =1+\frac{\frac1{24}}{\ln3-\frac{13}7\ln2}= 0.779$$

answered May 06 '20 at 19:50

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