Question

If $x_1$ and $x_2$ are two distinct solutions of the equation ${\log }_5({\log }_{64}|x|+(25{)}^x-\frac{1}{2})=2x,$ then

• A. $x_1=2x_2$
• B. $x_1+x_2=0$
• C. $x_1=3x_2$
• D. $x_1{x}_2=64$

Answer 1

The correct option is B

$x_1+x_2=0$

${\log }_5({\log }_{64}|x|+(25{)}^x-\frac{1}{2})=2x$

$\Rightarrow {\log }_{64}|x|+(25{)}^x-\frac{1}{2}=(25{)}^x$

$\Rightarrow {\log }_{64}|x|=\frac{1}{2}$

$\Rightarrow |x|=8$

$\Rightarrow x=-8,8$