Question

If $x^2+y^2=z^2$ then find the value of $\frac{1}{{\log }_{z-y}x}+\frac{1}{{\log }_{z+y}x}$.

Answer 1

Given,

$x^2+y^2=z^2$

or, $z^2-y^2=x^2$.......(1).

Now,

$\frac{1}{{\log }_{z-y}x}+\frac{1}{{\log }_{z+y}x}$

$=\frac{\log (z-y)}{\log x}+\frac{\log (z+y)}{\log x}$

$=\frac{\log (z^2-y^2)}{\log x}$

$=\frac{\log x^2}{\log x}$ [ Using (1)]

$=\frac{2\log x}{\log x}$

$=2$.