Fun with logarithms!

We have $\log_{2}{(2^{x}-1)}+x=\log_{4}{144} \\ \implies \log_{2}{(2^x-1)}+x=\log_{2^2}{12^2}=\log_{2}{12}\\ \implies x=\log_{2}{12 }-\log_{2}{(2^x-1)} \\ \implies x=\log_{2}{\left (\dfrac {12}{2^x-1}\right)} \\\implies 2^x=\dfrac {12}{2^x-1}$ Let $2^x$ be $a$ .The equation becomes : $a=\dfrac {12}{a-1} \implies a^2-a-12=0 \implies (a-4)(a+3)=0 \implies a=4,-3 \\\text{but}~2^x=-3~ \text{is undefined.}\\ \implies 2^x=4 \therefore \boxed {x=2}$

$\begin{aligned} \log_2(2^x-1) + x & = \log_4 144 \\ & = \frac {\log_2 144}{\log_2 4} \\ & = \frac {\log_2 2^4 \cdot 3^2}{\log_2 2^2} \\ & = \frac {4 + 2\log_2 3}2 \\ \implies \log_2 ({\color{#3D99F6}2^x-1}) + {\color{#D61F06}x} & = {\color{#D61F06}2} + \log_2 \color{#3D99F6}3 \end{aligned}$

Equating respective terms on both sides $\implies \begin{cases} \color{#3D99F6} 2^x - 1 = 3 & \implies 2^x = 4 & \implies x = \boxed 2 \\ \color{#D61F06} x = \boxed 2 \end{cases}$