# How to solve this? *log*

## ${\left({5}^{x + 1}\right)}^{2} = 0.2 \sqrt{{5}^{x}}$

Apr 18, 2016

$x = - 2$

#### Explanation:

Use the rule ${\left({a}^{b}\right)}^{c} = {a}^{b c}$ to simplify the left hand side of the equation. Also, write $\sqrt{{5}^{x}}$ as ${\left({5}^{x}\right)}^{\frac{1}{2}}$ and simplify it using the same rule, to give:

${5}^{2 x + 2} = 0.2 \left({5}^{\frac{1}{2} x}\right)$

Now, note that $0.2 = \frac{1}{5} = {5}^{-} 1$.

${5}^{2 x + 2} = {5}^{-} 1 \left({5}^{\frac{1}{2} x}\right)$

Simplify the right hand side using the rule ${a}^{b} \left({a}^{b}\right) = {a}^{b + c}$.

${5}^{2 x + 2} = {5}^{\frac{1}{2} x - 1}$

Now, since we have two exponential functions with the same base being equal, we also know that their exponents will be equal.

$2 x + 2 = \frac{1}{2} x - 1$

$\frac{3}{2} x = - 3$

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