Evaluate natural log of e^4+ natural log of e^-7

$\ln (e^{4}) + \ln (e^{- 7})$

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

$\ln (e^{4} e^{- 7})$

Multiply $e^{4}$ by $e^{- 7}$ by adding the exponents.

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

$\ln (e^{4 - 7})$

Subtract $7$ from $4$ .

$\ln (e^{- 3})$

Use logarithm rules to move $- 3$ out of the exponent.

$- 3 \ln (e)$

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

$- 3 \cdot 1$

Multiply $- 3$ by $1$ .

$- 3$