Question

Given $a^2+b^2=c^2$. Prove that ${\log }_{b+c}a+{\log }_{c-b}a=2{\log }_{b+c}a{\log }_{c-b}a$ for all $a>0,a\ne 1$

Answer 1

According to the definition of logarithms we have $c-b>0,c+b>0,c-b\ne 1,c+b\ne 1$
L.H.S $={\log }_{b+c}a+{\log }_{c-b}a$ ..........(1)
Using base change theorem in eqn$\left(1\right)$ we get
$=\frac{1}{{\log }_a(b+c)}+\frac{1}{{\log }_a(c-b)}$
$=\frac{{\log }_a(c-b)+{\log }_a(b+c)}{{\log }_a(c-b){\log }_a(b+c)}$
Using product law of logarithms to the Numerator, we get
$=\frac{{\log }_a(c^2-b^2)}{{\log }_a(c-b){\log }_a(b+c)}$
As $a^2+b^2=c^2;a^2=c^2-b^2$
$\frac{{\log }_a\left(a^2\right)}{{\log }_a(c-b){\log }_a(b+c)}=\frac{2{\log }_{a}a}{{\log }_a(c-b){\log }_a(b+c)}$
As ${\log }_{a}a=1$ we have
$\frac{2{\log }_{a}a}{{\log }_a(c-b){\log }_a(b+c)}=\frac{2}{{\log }_a(c-b){\log }_a(b+c)}$
$=2{\log }_{b+c}a{\log }_{c-b}a$ ...... (using base change theorem)