Question

Solve:
$\sqrt{5^0+\frac{3}{7}}=(0.49{)}^{2-3x}$

Answer 1

$\begin{array}{c}\sqrt{5^0+\frac{3}{7}}={\left(0.49\right)}^{2-3x}\\ \sqrt{1+\frac{3}{7}}={\left(\frac{49}{100}\right)}^{2-3x}\\ {\left(\frac{10}{7}\right)}^{\frac{1}{2}}={\left(\frac{7}{10}\right)}^{2\left(2-3x\right)}\\ {\left(\frac{7}{10}\right)}^{-\frac{1}{2}}={\left(\frac{7}{10}\right)}^{4-6x}\\ -\frac{1}{2}=4-6x\\ -1=8-12x\\ -1-8=+12x\\ +9=+12x\\ x=\frac{9}{12}\\ x=\frac{3}{4}\end{array}$