Question

Solve ${\log }_5\sqrt{7x-4}-\frac{1}{2}={\log }_5\sqrt{x+2}$

Answer 1

${\log }_5\sqrt{7x-4}-\frac{1}{2}={\log }_5\sqrt{x+2}$
${\log }_5\sqrt{7x-4}-{\log }_5\sqrt{x+2}=\frac{1}{2}$
${\log }_5\left(\frac{\sqrt{7x-4}}{\sqrt{x+2}}\right)=\frac{1}{2}$ $[\because {\log }_a\left(\frac{M}{N}\right)={\log }_{a}M-{\log }_{a}N]$
${\log }_5{\left(\frac{7x-4}{x+2}\right)}^{\frac{1}{2}}=\frac{1}{2}$
$\frac{1}{2}\left[{\log }_5\left(\frac{7x-4}{x+2}\right)\right]=\frac{1}{2}$ $[\because {\log }_a{M}^n=n{\log }_{a}M]$
${\log }_5\left(\frac{7x-4}{x+2}\right)=1$
$5^1=\frac{7x-4}{x+2}$ (exponential form)
Cross multiplying, $7x-4=5(x+2)$
$7x-4=5x+10⟹7x-5x=10+4$
$⟹2x=14$
$\therefore x=7$