# How do you solve 3^(2x+1) = 5?

Jun 30, 2017

I got:
$x = \frac{1}{2} \left[\ln \frac{5}{\ln} \left(3\right) - 1\right] = 0.23486$

#### Explanation:

We take the natural log of both sides:

$\ln {\left(3\right)}^{2 x + 1} = \ln \left(5\right)$

apply a property of logs and write:

$\left(2 x + 1\right) \ln \left(3\right) = \ln \left(5\right)$

rearrange:

$2 x + 1 = \ln \frac{5}{\ln} \left(3\right)$

$x = \frac{1}{2} \left[\ln \frac{5}{\ln} \left(3\right) - 1\right] = 0.23486$

Jun 30, 2017

$x \approx 0.232 \text{ to 3 dec. places}$

#### Explanation:

$\text{using the "color(blue)"law of logarithms}$

• logx^nhArrnlogx

${3}^{2 x + 1} = 5$

$\text{take ln (natural log) of both sides}$

$\Rightarrow \ln {3}^{2 x + 1} = \ln 5$

$\Rightarrow \left(2 x + 1\right) \ln 3 = \ln 5$

$\Rightarrow 2 x + 1 = \ln \frac{5}{\ln} 3 \leftarrow \text{ subtract 1 from both sides}$

$\Rightarrow 2 x = \left(\ln \frac{5}{\ln} 3\right) - 1 \leftarrow \text{ divide both sides by 2}$

$\Rightarrow x = \frac{1}{2} \left[\left(\ln \frac{5}{\ln} 3\right) - 1\right] \approx 0.232 \text{ 3 dec. places}$