# How do you solve 2^ { m + 1} + 9= 44?

Mar 3, 2018

$m = {\log}_{2} \left(35\right) - 1 \approx 4.13$

#### Explanation:

We start by subtracting $9$ from both sides:
${2}^{m + 1} + \cancel{9 - 9} = 44 - 9$

${2}^{m + 1} = 35$

Take ${\log}_{2}$ on both sides:
$\cancel{{\log}_{2}} \left({\cancel{2}}^{m + 1}\right) = {\log}_{2} \left(35\right)$

$m + 1 = {\log}_{2} \left(35\right)$

Subtract $1$ on both sides:
$m + \cancel{1 - 1} = {\log}_{2} \left(35\right) - 1$

$m = {\log}_{2} \left(35\right) - 1 \approx 4.13$

Mar 3, 2018

$m \approx 4.129$ (4sf)

#### Explanation:

${2}^{m + 1} + 9 = 44$

${2}^{m + 1} = 35$

In logarithm form, this is:

${\log}_{2} \left(35\right) = m + 1$

I remember this almost as keep 2 as the base and switch the other numbers.

$m = {\log}_{2} \left(35\right) - 1$

$m \approx 4.129$ (4sf)

Mar 3, 2018

$m = \frac{\log 35 - \log 2}{\log} 2$

#### Explanation:

${2}^{m + 1} + 9 = 44$

${2}^{m + 1} = 44 - 9 = 35$

$\log \left({2}^{m + 1}\right) = \log 35 \text{ }$ (taking the logarithm base $10$ on both sides)

$\log \left({2}^{m} \cdot 2\right) = \log 35$

$\log {2}^{m} + \log 2 = \log 35$

$\log {2}^{m} = \log 35 - \log 2$

$m \log 2 = \log 35 - \log 2$

$m = \frac{\log 35 - \log 2}{\log} 2$