Question
Solve for $x:\sqrt{\left(8^0+\frac{2}{3}\right)}=(0.6)^{2-3 x}$
✓
Answer
$ \sqrt{\left(8^0+\frac{2}{3}\right)}=(0.6)^{2-3 x}$
$\Rightarrow \left(1+\frac{2}{3}\right)^{\frac{1}{2}}=\left(\frac{6}{10}\right)^{2-3 x}$
$\Rightarrow \left(\frac{5}{3}\right)^{\frac{1}{2}}=\left(\frac{3}{5}\right)^{2-3 x}$
$\Rightarrow \left(\frac{3}{5}\right)^{-\frac{1}{2}}=\left(\frac{3}{5}\right)^{2-3 x}$
$\Rightarrow -\frac{1}{2}=2-3 x$
$\Rightarrow -1=4-6 x$
$\Rightarrow -5=-6 x$
$\Rightarrow x=\frac{5}{6}$
$\Rightarrow \left(1+\frac{2}{3}\right)^{\frac{1}{2}}=\left(\frac{6}{10}\right)^{2-3 x}$
$\Rightarrow \left(\frac{5}{3}\right)^{\frac{1}{2}}=\left(\frac{3}{5}\right)^{2-3 x}$
$\Rightarrow \left(\frac{3}{5}\right)^{-\frac{1}{2}}=\left(\frac{3}{5}\right)^{2-3 x}$
$\Rightarrow -\frac{1}{2}=2-3 x$
$\Rightarrow -1=4-6 x$
$\Rightarrow -5=-6 x$
$\Rightarrow x=\frac{5}{6}$
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