Proof the logarithmic identity $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$

Question

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.

Answer 1

Implement the formula:

1) $A^2-B^2=(A-B)(A+B)$

2) $\log_a xy=\log_a x+\log_a y$

3) $\log_a x^n=n\log_a x$

4) $\log_a a=1$

5) $\log_a b=\frac{1}{\log_b a}$

$a^2+b^2=c^2$

$a^2=c^2-b^2$/$\cdot\log_a$

$\log_a{a^2}=\log_a {(c^2-b^2)}$

$2\log_a a=\log_a{(c-b)(c+b)}$

$2=\log_a{(c+b)}+\log_a{(c-b)}$

$2=\frac{1}{\log_{c+b} a}+\frac{1}{\log_{c-b} a}$

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

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

Answer 2

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

Answer 3

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.