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

Nov 10, 2015

$x = \frac{\log \left(\frac{2}{25}\right)}{\log 5} = - 1 , 57$

#### Explanation:

From the laws of exponents we may write this as

${5}^{x} \cdot {5}^{2} = 2$

Now dividing throughout by ${5}^{2}$ yields

${5}^{x} = \frac{2}{25}$

Now taking the log on both sides and using laws of logs we can write it as

$x \log 5 = \log \left(\frac{2}{25}\right)$

Finally dividing throughout by $\log 5$ gives

$x = \frac{\log \left(\frac{2}{25}\right)}{\log 5} = - 1 , 57$

Nov 14, 2015

To undo the $x$ being raised to an exponent, take the log base 5 of both sides. This undoes the 5 on the left, leaving just $x + 2$. On the right is ${\log}_{\text{5}} 2$, which can be put into a calculator using the change of base formula, in which ${\log}_{\text{5}} 2 = \frac{\log 2}{\log 5}$.

Then, 2 is subtracted from both sides, leaving $x$ equal to $\frac{\log 2}{\log 5} - 2 \approx - 1.57$.