# How do you simplify ln e^(2x)?

Aug 25, 2016

$\ln {e}^{2 x} = 2 x$

#### Explanation:

As a Real valued function, $x \mapsto {e}^{x}$ is one to one from $\left(- \infty , \infty\right)$ onto $\left(0 , \infty\right)$.

As a result, for any $y \in \left(0 , \infty\right)$ there is a unique Real value $\ln y$ such that ${e}^{\ln y} = y$.

This is the definition of the Real natural logarithm.

If $t \in \left(- \infty , \infty\right)$ then $y = {e}^{t} \in \left(0 , \infty\right)$ and from the above definition:

${e}^{\ln \left({e}^{t}\right)} = {e}^{t}$

Since $x \mapsto {e}^{x}$ is one to one, we can deduce that for any Real value of $t$:

$\ln {e}^{t} = t$

In other words, $t \mapsto {e}^{t}$ and $t \mapsto \ln t$ are mutual inverses as Real valued functions.

So if $x$ is any Real value:

$\ln {e}^{2 x} = 2 x$

May 31, 2017

$2 x$

#### Explanation:

Using the property of logs:

$\log \left({a}^{b}\right) = b \log a$

We can see that:

$\ln \left({e}^{2 x}\right) = 2 x \ln e$

And since $\ln \left(e\right) = {\log}_{e} \left(e\right) = 1$,

$2 x \ln e = 2 x$

May 31, 2018

$2 x$

#### Explanation:

The key realization here is that $\ln x$ and ${e}^{x}$ are inverses of each other, which cancel each other out. So we essentially have

$\cancel{\ln} {\cancel{e}}^{2 x}$

which just leaves us with $\textcolor{b l u e}{2 x}$.

Hope this helps!