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

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


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Provided, of course, that $\,1\neq c-b>0\,$ . Have you tried some basic logarithmic properties? For example, $\,n\log_tk=\log_t(k^n)\,$ ... – DonAntonio Jan 8 '13 at 11:39
Should the relation be $c^2=a^2+b^2?$ – lab bhattacharjee Jan 8 '13 at 14:36

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 $log_{c-b}^{a}$

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write your equations on dollars, and for fractions use "\frac{}{}" as I did for you. And if you want to bring your equations in new line use double dollars. – AmirHosein SadeghiManesh Jan 8 '13 at 17:38

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 $A,B$.

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