# How do you solve 9^(n+10)+3=81?

Dec 18, 2016

$n = - \frac{7}{2}$

#### Explanation:

${9}^{n + 10} + 3 = 81$
${3}^{2 n + 10} + {3}^{1} = {3}^{4}$
$2 n + 10 + 1 = 4$
$2 n = 4 - 11$
$2 n = - 7$
$n = - \frac{7}{2}$
substitute$2 n = - 7$
$- 7 + 10 + 1 = 4$
$4 = 4$

Dec 23, 2016

$n \approx - 8.017$

#### Explanation:

${9}^{n + 10} + 3 = 81$

${9}^{n + 10} = 78 \text{ }$ the variable is in the index, so use logs

$\log {9}^{n + 10} = \log 78 \text{ } \leftarrow$take logs of both sides

$\left(n + 10\right) \log 9 = \log 78 \text{ } \leftarrow$ log law $\log {a}^{b} = b \log a$

$n + 10 = \log \frac{78}{\log} 9$

$n = \log \frac{78}{\log} 9 - 10 \text{ } \leftarrow$ use a calculator at this point

$n = - 8.017176$