Solve for d e^(2d)=5

$e^{2 d} = 5$

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

$\ln (e^{2 d}) = \ln (5)$

Expand the left side.

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

$2 d \ln (e) = \ln (5)$

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

$2 d \cdot 1 = \ln (5)$

Multiply $2$ by $1$ .

$2 d = \ln (5)$

Divide each term in $2 d = \ln (5)$ by $2$ and simplify.

Divide each term in $2 d = \ln (5)$ by $2$ .

$\frac{2 d}{2} = \frac{\ln (5)}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 d}{2} = \frac{\ln (5)}{2}$

Divide $d$ by $1$ .

$d = \frac{\ln (5)}{2}$

The result can be shown in multiple forms.

Exact Form:

$d = \frac{\ln (5)}{2}$

Decimal Form:

$d = 0.80471895 \dots$