Evaluate natural log of e^-9+ natural log of e^5

$\ln (e^{- 9}) + \ln (e^{5})$

Step 1

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

$\ln (e^{- 9} e^{5})$

Step 2

Multiply $e^{- 9}$ by $e^{5}$ by adding the exponents.

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

$\ln (e^{- 9 + 5})$

Add $- 9$ and $5$ .

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

Step 3

Use logarithm rules to move $- 4$ out of the exponent.

$- 4 \ln (e)$

Step 4

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

$- 4 \cdot 1$

Step 5

Multiply $- 4$ by $1$ .

$- 4$