Logs in an A.P.

Given that $\log_{10}{2}$ , $\log_{10}{(2^x-1)}$ and $\log_{10}{(2^x+3)}$ , then:

$\log_{10}{2} + \log_{10}{(2^x+3)} = 2\log_{10}{(2^x-1)}$

$\Rightarrow 2(2^x+3) = (2^x-1)^2$

$\quad 2(2^x)+6 = (2^x)^2-2(2^x)+1$

$\quad \Rightarrow (2^x)^2-4(2^x) -5 = 0$

$\quad \quad (2^x-5)(2^x+1) = 0$

$\quad \quad$ As $2^x \ne -1 \quad \Rightarrow 2^x = 5$

$\quad \quad \Rightarrow \log_2{2^x} = \log_2{5}$

$\quad \quad \Rightarrow x = \log_2{5} = \log_a{b}$

$\quad \quad \Rightarrow a = 2$ and $b = 5$

Therefore, $(a+b)^3 = 7^3 = \boxed{343}$

I think the problem is badly stated, since you can choose other (less natural) values for a,b. For example a=3, b=3^x will also work and will given a different answer.