Proof of $\log_{b+c} {a}+ \log_{c-b}{a}= 2\log_{b+c} {a} \cdot \log_{c-b}{a}$ when $a^2=b^2+c^2 $

Question

If $ a^2=b^2+c^2 $ prove that

$$\log_{b+c} {a}+ \log_{c-b}{a}= 2\log_{b+c} {a} \cdot \log_{c-b}{a}.$$

thanks

Answer 1

At first start from right hand side An put $$\frac{1}{log_a^{b+c}}$$ instead of $log_{b+c}^{a}$ and $$\frac{1}{log_a^{c-b}}$$ instead of $log_{c-b}^{a}$

Answer 2

Clearly, we need $b\pm c\ne 1,0$ as $\log_0x$ and $\log_1x$ are undefined for any real $x$.

Again if $a=1\iff \log_ma=0$ for real $m\ne 0,1$ either sides become $0$ for all real $b,c$ satisfying the 1st condition.

Now, $\log_{b+c} {a}+ \log_{c-b}{a}$ will be equal to $2\log_{b+c} {a} \cdot \log_{c-b}{a}$

if $$\frac{\log a}{\log (c+b)}+\frac{\log a}{\log (c-b)}=2\left(\frac{\log a}{\log (c+b)}\frac{\log a}{\log(c-b)}\right)$$ as $\log_CD=\frac{\log_nC}{\log_nD}$ where $D,n\ne1>0$

or if $$\log (c+b)+\log(c-b)=2\log a$$assuming $c\pm b\ne1$ cancelling $\log a$ as $\log a\ne0$

or if $$(c+b)(c-b)=a^2$$ as $\log A+\log B=\log AB$ for any real $A,B$.