Solve for n e^(5n-1)-40=14

$e^{5 n - 1} - 40 = 14$

Step 1

Move all terms not containing $n$ to the right side of the equation.

Add $40$ to both sides of the equation.

$e^{5 n - 1} = 14 + 40$

Add $14$ and $40$ .

$e^{5 n - 1} = 54$

Step 2

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

$\ln (e^{5 n - 1}) = \ln (54)$

Step 3

Expand the left side.

Expand $\ln (e^{5 n - 1})$ by moving $5 n - 1$ outside the logarithm.

$(5 n - 1) \ln (e) = \ln (54)$

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

$(5 n - 1) \cdot 1 = \ln (54)$

Multiply $5 n - 1$ by $1$ .

$5 n - 1 = \ln (54)$

Step 4

Add $1$ to both sides of the equation.

$5 n = \ln (54) + 1$

Step 5

Divide each term in $5 n = \ln (54) + 1$ by $5$ and simplify.

Divide each term in $5 n = \ln (54) + 1$ by $5$ .

$\frac{5 n}{5} = \frac{\ln (54)}{5} + \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 n}{5} = \frac{\ln (54)}{5} + \frac{1}{5}$

Divide $n$ by $1$ .

$n = \frac{\ln (54)}{5} + \frac{1}{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$n = \frac{\ln (54)}{5} + \frac{1}{5}$

Decimal Form:

$n = 0.99779680 \dots$