1. ## Logarithm equation

Solve $\log_{5}(x + 1) + 3\log_{2} 0.5 = \log_{25} (2x)^2$
So these are the steps I have done thus far:
$\log_{5} (x + 1) + \log_{2} \frac{1}{8} = \frac{\log_{5} (2x)^2}{\log_{5} 25}$
$\log_{5} (x + 1) + \log_{2} \frac{1}{8} = \frac{1}{2} (\log_{5} (2x)^2)$
$\log_{5}(x + 1) + \log_{2} \frac{1}{8} = \log_{5} (2x)$
$\log_{2} \frac{1}{8} = \log_{5}(2x^2 + 2x)$
$\frac{\log_{5} \frac{1}{8}} {\log_{5} 2} = \log_{5} (2x^2 + 2x)$
$\log_{5} \frac{1}{8} = [\log_{5} (2x^2 + 2x)][\log_{5}2]$

I feel as if I'm nowhere close to getting the final answer which is supposed to be $x = \frac{1}{249}$. Any help and would really be appreciated and thank you in advance.

2. Notice that $3\log_2(0.5)=-3$.

3. oh!! I can find the answer now
please forgive my lack of latex:

Thank you very much for the pointer!