Question

If ${\log }_{x}49-{\log }_{x}7+{\log }_x\frac{1}{343}+2=0$, then the value of $x$ is equal to

• A. $7^2$
• B. $7^1$
• C. $7^3$
• D. $7^{-1}$

Answer 1

The correct option is B $7^1$

${\log }_x{7}^2-{\log }_{x}7+{\log }_x\frac{1}{7^3}+2=0{\log }_x\frac{7^2}{7}\times \frac{1}{7^3}=-2{\log }_x{7}^{-2}=-27^{-2}=x^{-2}\Rightarrow x=7$