Prove: $\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$, where $a^2 + b^2 = c^2$ Could you help me proving this?

$$\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$$
  where $c$ is the length of the hypotenuse of a right triangle, and $a$ and $b$ are the lengths of the other sides.

I tried that but not sure if it helps and what to do next:
$$\frac{\log\left(a\right)}{\log\left(c+b\right)}+\frac{\log\left(a\right)}{\log\left(c-b\right)}=\frac{2\log^2\left(a\right)}{\log\left(c+b\right)\log\left(c-b\right)}$$
Thank you in advance! 
 A: We have $$a^2+b^2=c^2$$ so that $$(c+b)(c-b)=c^2-b^2=a^2.$$ Take the logarithm of both sides $$\log(c+b)+\log(c-b)=2\log(a).$$
Divide by $\log(c+b)$:
$$\frac{\log(c+b)}{\log(c+b)}+\frac{\log(c-b)}{\log(c+b)}=2\frac{\log(a)}{\log(c+b)}$$
Simplify, using the fact that $\frac{\log(x)}{\log(y)}=\log_y(x)$:
$$1+\log_{c+b}(c-b)=2\log_{c+b}(a).$$
Multiply by $\log_{c-b}(a)$:
$$\log_{c-b}(a)+\log_{c-b}(a)\log_{c+b}(c-b)=2\log_{c+b}(a)\log_{c-b}(a).$$
Finally,  $$\log_{c-b}(a)\log_{c+b}(c-b)=\frac{\log(a)}{\log(c-b)}\frac{\log(c-b)}{\log(c+b)}=\frac{\log(a)}{\log(c+b)}=\log_{c+b}(a).$$
A: The change of base formula for logarithms says that
$$
\log_xy=\frac{1}{\log_yx}
$$
assuming $x$ and $y$ positive and different from $1$. Thus, assuming $a\ne1$, $c-b\ne1$ and $c+b\ne1$, we have
\begin{align}
\log_{c+b}a+\log_{c-b}a
&=
\frac{1}{\log_a(c+b)}+\frac{1}{\log_a(c-b)}\\[6px]
&=
\frac{\log_a(c-b)+\log_a(c+b)}{\log_a(c+b)\cdot\log_a(c-b)}\\[6px]
&=
\frac{\log_a\bigl((c-b)(c+b)\bigr)}{\log_a(c+b)\cdot\log_a(c-b)}\\[6px]
&=
\frac{\log_a(c^2-b^2)}{\log_a(c+b)\cdot\log_a(c-b)}
\end{align}
Since $c^2-b^2=a^2$, the numerator is $\log_aa^2=2$ and, applying again the change of base,
$$
\log_{c+b}a+\log_{c-b}a=2\log_{c+b}a\cdot\log_{c-b}a
$$
If $a=1$ the equality is obvious. If either $c+b=1$ or $c-b=1$, the equality doesn't make sense.
