# How do you solve 5^(x+2)=4?

Jul 25, 2016

$x = \frac{2 \cdot \left(\log 2 - \log 5\right)}{\log} 5$

#### Explanation:

One of the logarithm rules one should keep in mind for this problem:

$\log {a}^{b} = b \cdot \log a$

Apply logarithm on both sides

$\log \left({5}^{x + 2}\right) = \log 4$

$\implies \left(x + 2\right) \cdot \log 5 = \log 4$

$\implies x + 2 = \log \frac{4}{\log} 5$

Now it's just a matter of simplification:

$\implies x = \log \frac{{2}^{2}}{\log} 5 - 2$

$\implies x = \frac{2 \cdot \log 2}{\log} 5 - 2$

$\implies x = \frac{2 \cdot \log 2 - 2 \log 5}{\log} 5$

$\mathmr{and} , x = \frac{2 \cdot \left(\log 2 - \log 5\right)}{\log} 5$