# Question #6daa9

Apr 5, 2016

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

#### Explanation:

Take the natural logarithm on both sides

$\ln \left({e}^{3 x} \cdot {e}^{2 x - 1}\right) = \ln \left(2 e\right)$

From the identity

$\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$

We can simplify the above equation as

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

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

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

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

Apr 5, 2016

A slightly different approach

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

#### Explanation:

Given:$\text{ } {e}^{3 x} \times {e}^{2 x - 1} = 2 e$

Compare to ${10}^{2} \times {10}^{1} = {10}^{3} = {10}^{2 + 1}$

Using the above method we have;

$\text{ "e^(3x) xx e^(2x-1)=2e" "->" } {e}^{3 x + 2 x - 1} = 2 e$

Divide both sides by $e$

${e}^{3 x + 2 x - 2} = 2$

$\implies {e}^{5 x - 2} = 2$

Take logs of both sides

$\left(5 x - 2\right) \ln \left(e\right) = \ln \left(2\right)$

But $\ln \left(e\right) = 1$

$5 x - 2 = \ln \left(2\right)$

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