# How do you solve 20^(9-4n)=5?

Jul 20, 2017

$n = \frac{9 - {\log}_{20} 5}{4}$

#### Explanation:

We know that :

${x}^{y} = z \iff y = {\log}_{x} z$

so :

${20}^{9 - 4 n} = 5 \iff 9 - 4 n = {\log}_{20} 5 \iff 4 n = 9 - {\log}_{20} 5 \iff$

$n = \frac{9 - {\log}_{20} 5}{4}$

Jul 20, 2017

$n = \frac{9}{4} - \frac{1}{4} \left(\log \frac{5}{\log} 20\right)$

$n = 2.116$

#### Explanation:

In a question like this you first have to decide whether you will solve it as an exponential equation, or use logs.

$5$ is certainly not a power of $20$, so logs are indicated.

${20}^{9 - 4 n} = 5 \text{ } \leftarrow$ find log of both sides

$\log {20}^{\textcolor{b l u e}{9 - 4 n}} = \log 5 \text{ } \leftarrow$ apply the power law

$\textcolor{b l u e}{\left(9 - 4 n\right)} \log 20 = \log 5 \text{ } \leftarrow$ isolate the factor with $n$

$9 - 4 n = \log \frac{5}{\log} 20 \text{ } \leftarrow$ re-arrange to get $4 n$ positive

$9 - \log \frac{5}{\log} 20 = 4 n \text{ } \leftarrow$ divide by 4

$\frac{1}{4} \left(9 - \log \frac{5}{\log} 20\right) = n$

$n = \frac{9}{4} - \frac{1}{4} \left(\log \frac{5}{\log} 20\right)$

$n = 2.116$ (to 3 d.p.)