# If 5^x = 3, what does 5^-(2x) equal?

Jun 28, 2017

$\frac{1}{9}$

#### Explanation:

1. First, take the log (base 10) of both sides.

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

2. When you have an exponent inside of a log, you can take that exponent out and multiply it by the rest of the log.

$\log \left({x}^{y}\right) = y \cdot \log \left(x\right)$

This is known as the power rule of logarithms.

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

3. Now divide both sides by $\log \left(5\right)$ to find $x$.

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

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

4. Now, plug $x$ (in terms of the logs) into the second expression.

5^(-(2* log(3)/log(5))

Let's evaluate the power first. (I used a calculator here.)

$- \left(2 \cdot \log \frac{3}{\log} \left(5\right)\right) \approx - 1.365212389$

${5}^{-} 1.365212389 = \overline{.1} = \frac{1}{9}$

So ${5}^{- 2 x} = \frac{1}{9}$.

Hope this helps! :)