Question

Solve for $x$, if
$log{x}^{49}-{log}_x^7+{log}_x\frac{1}{343}+2=0$.

Answer 1

${\log }_{x}49-{\log }_{x}7+{\log }_x\left(\frac{1}{343}\right)+2=0$

$\Rightarrow {\log }_x(\frac{49}{7}\times \frac{1}{343})=-2$

$\Rightarrow {\log }_x(\frac{7^2}{7}\times \frac{1}{7^3})=-2$

$\Rightarrow {\log }_x\left(7^{2-1-3}\right)=-2$

$\Rightarrow {\log }_x\left(7^{-2}\right)=-2$

$\Rightarrow -2{\log }_{x}7=-2$

$\Rightarrow {\log }_{x}7=1$

$\therefore x=7$