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

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

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

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

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

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

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

Add $6$ and $12$ .

$\ln (e^{18})$

Use logarithm rules to move $18$ out of the exponent.

$18 \ln (e)$

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

$18 \cdot 1$

Multiply $18$ by $1$ .

$18$