Question

If ${\log }_{x}49-{\log }_{x}7+{\log }_x\frac{1}{343}+2=0$, then

• A. $x=7$
• B. $x=3$
• C. $x=12$
• D. $x=16$

Answer 1

The correct option is A $x=7$

${\log }_{x}49-{\log }_{x}7+{\log }_x\frac{1}{343}+2=0{\log }_x\left(\frac{49}{7\times 343}\right)=-2({\log }_{a}b+{\log }_{a}c=lo{g}_{a}bc){\log }_x\frac{1}{49}=-2\frac{1}{49}=x^{-2}So,x=7$