Question

Solve:
$\sqrt[3]{34^0+\frac{3}{5}}={\left(\frac{5}{8}\right)}^{1-2x}$

Answer 1

$3\sqrt{4^{\circ }+\frac{3}{5}}={\left(\frac{5}{8}\right)}^{1-2x}$

Taking both side we get

$\begin{array}{cc}{4}^{\circ }+\frac{3}{5}={\left[{\left(\frac{5}{8}\right)}^{1-2x}\right]}^3& \\ \frac{1}{1}+\frac{3}{5}={\left(\frac{5}{8}\right)}^{3-6x}& \left[{\left(a^m\right)}^n=a^{mn}\right]\\ {\left(\frac{8}{5}\right)}^1={\left(\frac{5}{8}\right)}^{3-6x}& \\ {\left(\frac{5}{8}\right)}^{-1}={\left(\frac{5}{8}\right)}^{3-6x}& \left[Ifbaseissavethenequatethepower\right]\\ -1=3-6x& \\ -1-3=-6x& \\ 4=6x& \\ \frac{4}{6}=x& \\ x=\frac{2}{3}& \end{array}$