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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.

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A good approach: $$\log_2 (2^x-1)+x=\log_4 (144)$$ $$\log_2 (2^x-1)+x=\log_2 (12)$$ $$x=\log_2 (12)-\log_2 (2^x-1)$$ $$x=\log_2\frac{12}{2^x-1}$$ $$2^x=\frac{12}{2^x-1}$$ Say $2^x=a$ $$a^2-a-12=0$$ $$(a-4)(a+3)=0$$ $$a=4,-3$$

But $2^x$ cannot be negative.

So, $$2^x=4=2^2$$ i.e. $$x=2$$

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  • $\begingroup$ Thanks Aniket. You're my hero! $\endgroup$ – RubenDefour Oct 31 '15 at 10:24
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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 $$2^x=4\iff 2^x=2^2$$ $$\color{red}{x=2}$$

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