Solving a Complex Exponential Function

Question

I recently had a quiz that asked a question similar to the following "Solve $3^{1-2x}=4^x\;$". What I am wondering is how you can go from "$x\log_3(4)=1-2x$" to the solution on problems like this. There are other variations on the equation but they all are very similar.

Answer 1

$$3^{1 - 2x} = 4^x$$ Take the log of any base on both sides. For simplicity, I will just use $\log_{10}$. $$\log(3^{1 - 2x}) = \log(4^x)$$ $$(1 - 2x)\log(3) = x\log(4)$$ $$\log(3) - x[2\log(3) - \log(4)] = 0$$ $$x = \frac{\log(3)}{2\log(3) - \log(4)}$$ The base I used had no effect on the answer. So, had I used base 3, we would find: $$x = \frac{1}{2 - \log_{3}(4)}$$