Question

Find the value of x for which ${\left(\frac{3}{5}\right)}^{2x}$=$\left(\frac{9}{25}\right)$.

Answer 1

We have ${\left(\frac{3}{5}\right)}^{2x}=\left(\frac{9}{25}\right)$

take log on both side

$log{\left(\frac{3}{5}\right)}^{2x}=log\left(\frac{9}{25}\right)$

$\left(2x\right)log\left(\frac{3}{5}\right)=log{\left(\frac{3}{5}\right)}^2$

$\left(2x\right)=\left(log\left(\frac{3}{5}\right)\right)=\left(2\right)log\left(\frac{3}{5}\right)$

$2x=2$

$x=1$