<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] $$\log_{c+b}a+\log_{c-b}a=\frac{\log a}{\log (b+c)}+\frac{\log a}{\log (c-b)}$$ $$=(\log a)(\frac{1}{\log (b+c)}+\frac{1}{\log (c-b)})$$ $$=(\log a)(\frac{\log(c+b)+\log(c-b)}{\log(c+b)\log(c-b)})$$ $$=(\log a)(\frac{\log(c^2-b^2)}{\log(c+b)\log(c-b)})$$ $$=(\log a)(\frac{\log(a^2)}{\log(c+b)\log(c-b)})$$ $$=2(\frac{\log a}{\log(c+b)})(\frac{\log a}{\log(c-b)})$$ $$=2\log_{c+b}a\cdot\log_{c-b} a$$
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