Question

Given $a^2+b^2=c^2$ & $a>0;b>0;c>0,c-b\ne 1,c+b\ne 1,$ Then : ${\log }_{c+b}a+{\log }_{c-b}a$ is equal to

• A. ${\log }_{c+b}a.{\log }_{c-b}a$
• B. $2.{\log }_{c+b}a.{\log }_{c-b}a$
• C. $\frac{{\log }_{c+b}a.{\log }_{c-b}a}{2}$
• D. $4.{\log }_{c+b}a.{\log }_{c-b}a$

Answer 1

The correct option is B $2.{\log }_{c+b}a.{\log }_{c-b}a$

$a^2+b^2=c^2\Rightarrow c^2-b^2=a^2$

Now,

${\log }_{c+b}a+{\log }_{c-b}a=\frac{1}{{\log }_{a}c+b}+\frac{1}{{\log }_{a}c-b}$

$=\frac{{\log }_{a}c-b+{\log }_{a}c+b}{\left({\log }_{a}c+b\right)\left({\log }_{a}c-b\right)}$

$=\frac{{\log }_a{c}^2-b^2}{\left({\log }_{a}c+b\right)\left({\log }_{a}c-b\right)}$
$=\frac{2}{\left({\log }_{a}c+b\right)\left({\log }_{a}c-b\right)}$

$=2{\log }_{c-b}a{\log }_{c+b}a\dots (\because {\log }_{a}b=\frac{1}{{\log }_{b}a})$

${\log }_{c+b}a+{\log }_{c-b}a=2{\log }_{c-b}a{\log }_{c+b}a$