Proof the logarithmic identity $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$ Please help me proof $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$, for $a,b,c>0$ and $a^2+b^2=c^2$. Thanks.

Implement the formula:
1) $A^2-B^2=(A-B)(A+B)$
2) $\log_a xy=\log_a x+\log_a y$
3) $\log_a x^n=n\log_a x$
4) $\log_a a=1$
5) $\log_a b=\frac{1}{\log_b a}$
$a^2+b^2=c^2$
$a^2=c^2-b^2$/$\cdot\log_a$
$\log_a{a^2}=\log_a {(c^2-b^2)}$
$2\log_a a=\log_a{(c-b)(c+b)}$
$2=\log_a{(c+b)}+\log_a{(c-b)}$
$2=\frac{1}{\log_{c+b} a}+\frac{1}{\log_{c-b} a}$
$2=\frac{\log_{c-b} a+\log_{c+b} a}{\log_{c-b} a\cdot\log_{c+b} a}$
$\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$
