Solve for x e^(4x-2)>=16

$e^{4 x - 2} \geq 16$

Step 1

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

$\ln (e^{4 x - 2}) \geq \ln (16)$

Step 2

Expand the left side.

Expand $\ln (e^{4 x - 2})$ by moving $4 x - 2$ outside the logarithm.

$(4 x - 2) \ln (e) \geq \ln (16)$

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

$(4 x - 2) \cdot 1 \geq \ln (16)$

Multiply $4 x - 2$ by $1$ .

$4 x - 2 \geq \ln (16)$

Step 3

Add $2$ to both sides of the inequality.

$4 x \geq \ln (16) + 2$

Step 4

Divide each term in $4 x \geq \ln (16) + 2$ by $4$ and simplify.

Divide each term in $4 x \geq \ln (16) + 2$ by $4$ .

$\frac{4 x}{4} \geq \frac{\ln (16)}{4} + \frac{2}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \geq \frac{\ln (16)}{4} + \frac{2}{4}$

Divide $x$ by $1$ .

$x \geq \frac{\ln (16)}{4} + \frac{2}{4}$

Simplify the right side.

Simplify each term.

Rewrite $\ln (16)$ as $\ln (2^{4})$ .

$x \geq \frac{\ln (2^{4})}{4} + \frac{2}{4}$

Expand $\ln (2^{4})$ by moving $4$ outside the logarithm.

$x \geq \frac{4 \ln (2)}{4} + \frac{2}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$x \geq \frac{4 \ln (2)}{4} + \frac{2}{4}$

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

$x \geq \ln (2) + \frac{2}{4}$

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$x \geq \ln (2) + \frac{2 (1)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$x \geq \ln (2) + \frac{2 \cdot 1}{2 \cdot 2}$

Cancel the common factor.

$x \geq \ln (2) + \frac{2 \cdot 1}{2 \cdot 2}$

Rewrite the expression.

$x \geq \ln (2) + \frac{1}{2}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \geq \ln (2) + \frac{1}{2}$

Interval Notation:

$[\ln (2) + \frac{1}{2}, \infty)$

Step 6

Solve for x e^(4x-2)>=16

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