Evaluate 8(e)^(2x+1)=4

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

Step 1

Divide each term in $8 e^{2 x + 1} = 4$ by $8$ and simplify.

Divide each term in $8 e^{2 x + 1} = 4$ by $8$ .

$\frac{8 e^{2 x + 1}}{8} = \frac{4}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 e^{2 x + 1}}{8} = \frac{4}{8}$

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

$e^{2 x + 1} = \frac{4}{8}$

Simplify the right side.

Cancel the common factor of $4$ and $8$ .

Factor $4$ out of $4$ .

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

Cancel the common factors.

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 2

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

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

Step 3

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 $1$ .

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

Step 4

Subtract $1$ from both sides of the equation.

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

Step 5

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.

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.84657359 \dots$

Evaluate 8(e)^(2x+1)=4

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