Solve the Inequality for x e^(3x+4)>11

$e^{3 x + 4} > 11$

Step 1

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

$\ln (e^{3 x + 4}) > \ln (11)$

Step 2

Expand the left side.

Expand $\ln (e^{3 x + 4})$ by moving $3 x + 4$ outside the logarithm.

$(3 x + 4) \ln (e) > \ln (11)$

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

$(3 x + 4) \cdot 1 > \ln (11)$

Multiply $3 x + 4$ by $1$ .

$3 x + 4 > \ln (11)$

Step 3

Subtract $4$ from both sides of the inequality.

$3 x > \ln (11) - 4$

Step 4

Divide each term in $3 x > \ln (11) - 4$ by $3$ and simplify.

Divide each term in $3 x > \ln (11) - 4$ by $3$ .

$\frac{3 x}{3} > \frac{\ln (11)}{3} + \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} > \frac{\ln (11)}{3} + \frac{- 4}{3}$

Divide $x$ by $1$ .

$x > \frac{\ln (11)}{3} + \frac{- 4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x > \frac{\ln (11)}{3} - \frac{4}{3}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x > \frac{\ln (11)}{3} - \frac{4}{3}$

Interval Notation:

$(\frac{\ln (11)}{3} - \frac{4}{3}, \infty)$

Step 6