# How do you solve 2lnx=1?

Mar 18, 2018

$x = {e}^{\frac{1}{2}}$

#### Explanation:

Let's do PEMDAS backwards.

We don't have any addition or subtraction, so we can't really do anything there.

We have multiplication that we can undo to isolate the ln(x):
$2 \ln x = 1$
$\ln x = \frac{1}{2}$

Now that the ln(x) is isolated, we can exponentiate:
$\ln x = \frac{1}{2} \implies {e}^{\ln x} = {e}^{\frac{1}{2}} \implies x = {e}^{\frac{1}{2}}$

Mar 18, 2018

see below.

#### Explanation:

1) Divide each term by $2$
$\ln \left(x\right) = \frac{1}{2}$

2) In solving $x$, we rewrite equation in logarithms form.

${e}^{\ln \left(x\right)} = {e}^{\frac{1}{2}}$

3) Exponential and base-e log are inverse function.
$x = {e}^{\frac{1}{2}}$

4) The value of $x$ is ${e}^{\frac{1}{2}}$ or $1.6487 \left(4 d . p .\right)$