# How do you solve ln((e^(4x+3))/e)=1?

Dec 15, 2015

$x = - \frac{1}{4}$

#### Explanation:

Use the following logarithmic law first:

$\ln \left(\frac{a}{b}\right) = \ln \left(a\right) - \ln \left(b\right)$

$\ln \left({e}^{4 x + 3} / e\right) = 1$

$\iff \ln \left({e}^{4 x + 3}\right) - \ln \left(e\right) = 1$

As next, you need to use the property that $\ln x$ and ${e}^{x}$ are inverse functions which means that $\ln \left({e}^{x}\right) = x$ and ${e}^{\ln x} = x$ always hold.

Thus, $\ln$ and $e$ eliminate each other in your equation, and you will get:

$\iff \left(4 x + 3\right) - 1 = 1$

The solution of this equation is

$x = - \frac{1}{4}$

As ${e}^{x}$ is always positive for any value of $x \in \mathbb{R}$, and thus the logarithmic expression is defined for any $x \in \mathbb{R}$, this is your solution.