Find All Complex Solutions 8(e)^(2x+1)=4

$8 {(e)}^{2 x + 1} = 4$

Subtract $4$ from both sides of the equation.

$8 {(e)}^{2 x + 1} - 4 = 0$

Remove parentheses.

$8 e^{2 x + 1} - 4 = 0$

Factor $4$ out of $8 e^{2 x + 1} - 4$ .

Factor $4$ out of $8 e^{2 x + 1}$ .

$4 (2 e^{2 x + 1}) - 4 = 0$

Factor $4$ out of $- 4$ .

$4 (2 e^{2 x + 1}) + 4 (- 1) = 0$

Factor $4$ out of $4 (2 e^{2 x + 1}) + 4 (- 1)$ .

$4 (2 e^{2 x + 1} - 1) = 0$

Divide each term in $4 (2 e^{2 x + 1} - 1) = 0$ by $4$ and simplify.

Divide each term in $4 (2 e^{2 x + 1} - 1) = 0$ by $4$ .

$\frac{4 (2 e^{2 x + 1} - 1)}{4} = \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 (2 e^{2 x + 1} - 1)}{4} = \frac{0}{4}$

Divide $2 e^{2 x + 1} - 1$ by $1$ .

$2 e^{2 x + 1} - 1 = \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

$2 e^{2 x + 1} - 1 = 0$

Add $1$ to both sides of the equation.

$2 e^{2 x + 1} = 1$

Divide each term in $2 e^{2 x + 1} = 1$ by $2$ and simplify.

Divide each term in $2 e^{2 x + 1} = 1$ by $2$ .

$\frac{2 e^{2 x + 1}}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 e^{2 x + 1}}{2} = \frac{1}{2}$

Divide $e^{2 x + 1}$ by $1$ .

$e^{2 x + 1} = \frac{1}{2}$

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

$\ln (e^{2 x + 1}) = \ln (\frac{1}{2})$

Expand the left side.

Expand $\ln (e^{2 x + 1})$ by moving $2 x + 1$ outside the logarithm.

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

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

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

Multiply $2 x + 1$ by $1$ .

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

Subtract $1$ from both sides of the equation.

$2 x = \ln (\frac{1}{2}) - 1$

Divide each term in $2 x = \ln (\frac{1}{2}) - 1$ by $2$ and simplify.

Divide each term in $2 x = \ln (\frac{1}{2}) - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

$x = \frac{\ln (\frac{1}{2})}{2} - \frac{1}{2}$

Find All Complex Solutions 8(e)^(2x+1)=4

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