Evaluate natural log of e^6+ natural log of e^-12

$\ln (e^{6}) + \ln (e^{- 12})$

Step 1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (e^{6} e^{- 12})$

Step 2

Multiply $e^{6}$ by $e^{- 12}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\ln (e^{6 - 12})$

Subtract $12$ from $6$ .

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

Step 3

Use logarithm rules to move $- 6$ out of the exponent.

$- 6 \ln (e)$

Step 4

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

$- 6 \cdot 1$

Step 5

Multiply $- 6$ by $1$ .

$- 6$