Question

If $x^2$ + $y^2$ = $z^2$. Then ${\log }_{y+z}x$ + ${\log }_{z-y}x$ = ?

• A. .
• B. 4.
• C. 2.
• D. 8.

Answer 1

The correct option is C

2.

Given $x^2$ + $y^2$ = $z^2$ $\Rightarrow x^2=z^2-y^2$
$=\frac{1}{lo{g}_x(y+z}+\frac{1}{lo{g}_x(z-y)}$
$=\frac{lo{g}_x(z-y)+lo{g}_x(z+y)}{lo{g}_x(y+z)lo{g}_x(z-y)}$
$=\frac{lo{g}_x(z^2-y^2)}{lo{g}_x(y+z).lo{g}_x(z-y)}=lo{g}_x{x}^2\times lo{g}_{(y+z)}xlo{g}_{(z-y)}x$
$=2lo{g}_{(y+z)}xlo{g}_{(z-y)}x$