<s>[INST] 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.
 [/INST] Note that $\log_xy=\displaystyle\frac{\ln y}{\ln x}$, here $\ln$ is the natural log. Therefore, we have
$$\log_{b+c} a+\log_{c-b} a=\frac{\ln a}{\ln(b+c)}+\frac{\ln a}{\ln(c-b)}
=\ln a\left(\frac{\ln(b+c)+\ln(c-b)}{\ln(b+c)\ln(c-b)}\right)$$
$$=\ln a\cdot\frac{\ln[(b+c)(c-b)]}{\ln(b+c)\ln(c-b)}
=\frac{\ln a\ln(c^2-b^2)}{\ln(b+c)\ln(c-b)}=\frac{\ln a\ln(a^2)}{\ln(b+c)\ln(c-b)}$$
where we have used the assumption $a^2+b^2=c^2$ in the last equality. 
Hence, we have
$$\log_{b+c} a+\log_{c-b} a=\frac{\ln a\ln(a^2)}{\ln(b+c)\ln(c-b)}
=\frac{2\ln a\ln a}{\ln(b+c)\ln(c-b)}$$
$$=2\cdot\frac{\ln a}{\ln(b+c)}\cdot\frac{\ln a}{\ln(c-b)}=2\log_{b+c}a\cdot\log_{c-b}a$$
as required. 
</s>