TITLE: Solving a logarithmic equation $\log_2 (2^x-1)+x=\log_4 (144)$
QUESTION [7 upvotes]: I need to solve this:
$$\log_2 (2^x-1)+x=\log_4 (144)$$
I can work out that $x=\log_2 (2^x)$ and $\log_4 (144)=log_2(12)$
but I'm stuck after that.

REPLY [2 votes]: Notice, logarithm formula: $\color{blue}{\log_{a^m}(b^n)=\frac{n}{m}\log_a(b)}$,  $$\log_2(2^x-1)+x=\log_4(144)$$
$$\log_2(2^x-1)+\log_2(2^x)=\log_{2^2}(12^2)$$
$$\log_2(2^x(2^x-1))=\frac{2}{2}\log_{2}(12)=\log_2(12)$$
since, logarithm is one-to-one function hence, comparing the numbers on the same base $2$ on both the sides, one should get $$2^x(2^x-1)=12$$
$$(2^x)^2-2^x-12=0$$
$$(2^x+3)(2^x-4)=0$$ $$\implies 2^x=-3, 4$$ but $2^x>0\ \ \ \forall \ \ \ x\in R$, hence one should have $$2^x=4\iff 2^x=2^2$$ $$\color{red}{x=2}$$