Evaluate e^(3x+6)=8

$e^{3 x + 6} = 8$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{3 x + 6}) = \ln (8)$

Step 2

Expand the left side.

Expand $\ln (e^{3 x + 6})$ by moving $3 x + 6$ outside the logarithm.

$(3 x + 6) \ln (e) = \ln (8)$

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

$(3 x + 6) \cdot 1 = \ln (8)$

Multiply $3 x + 6$ by $1$ .

$3 x + 6 = \ln (8)$

Step 3

Subtract $6$ from both sides of the equation.

$3 x = \ln (8) - 6$

Step 4

Divide each term in $3 x = \ln (8) - 6$ by $3$ and simplify.

Divide each term in $3 x = \ln (8) - 6$ by $3$ .

$\frac{3 x}{3} = \frac{\ln (8)}{3} + \frac{- 6}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{\ln (8)}{3} + \frac{- 6}{3}$

Divide $x$ by $1$ .

$x = \frac{\ln (8)}{3} + \frac{- 6}{3}$

Simplify the right side.

Simplify each term.

Rewrite $\ln (8)$ as $\ln (2^{3})$ .

$x = \frac{\ln (2^{3})}{3} + \frac{- 6}{3}$

Expand $\ln (2^{3})$ by moving $3$ outside the logarithm.

$x = \frac{3 \ln (2)}{3} + \frac{- 6}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x = \frac{3 \ln (2)}{3} + \frac{- 6}{3}$

Divide $\ln (2)$ by $1$ .

$x = \ln (2) + \frac{- 6}{3}$

Divide $- 6$ by $3$ .

$x = \ln (2) - 2$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \ln (2) - 2$

Decimal Form:

$x = - 1.30685281 \dots$