Evaluate natural log of 6/(e^2)

$\ln (\frac{6}{e^{2}})$

Step 1

Move $e^{2}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$\ln (6 e^{- 2})$

Step 2

Rewrite $\ln (6 e^{- 2})$ as $\ln (6) + \ln (e^{- 2})$ .

$\ln (6) + \ln (e^{- 2})$

Step 3

Use logarithm rules to move $- 2$ out of the exponent.

$\ln (6) - 2 \ln (e)$

Step 4

The natural logarithm of $e$ is $1$ .

$\ln (6) - 2 \cdot 1$

Step 5

Multiply $- 2$ by $1$ .

$\ln (6) - 2$

Step 6

Simplify each term.

Rewrite $\ln (6)$ as $\ln (2 (3))$ .

$\ln (2 (3)) - 2$

Rewrite $\ln (2 (3))$ as $\ln (2) + \ln (3)$ .

$\ln (2) + \ln (3) - 2$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\ln (2) + \ln (3) - 2$

Decimal Form:

$- 0.20824053 \dots$