Solve for x e^(1-5x)=793

$e^{1 - 5 x} = 793$

Step 1

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

$\ln (e^{1 - 5 x}) = \ln (793)$

Step 2

Expand the left side.

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

$(1 - 5 x) \ln (e) = \ln (793)$

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

$(1 - 5 x) \cdot 1 = \ln (793)$

Multiply $1 - 5 x$ by $1$ .

$1 - 5 x = \ln (793)$

Step 3

Subtract $1$ from both sides of the equation.

$- 5 x = \ln (793) - 1$

Step 4

Divide each term in $- 5 x = \ln (793) - 1$ by $- 5$ and simplify.

Divide each term in $- 5 x = \ln (793) - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{\ln (793)}{- 5} + \frac{- 1}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{\ln (793)}{- 5} + \frac{- 1}{- 5}$

Divide $x$ by $1$ .

$x = \frac{\ln (793)}{- 5} + \frac{- 1}{- 5}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$x = - \frac{\ln (793)}{5} + \frac{- 1}{- 5}$

Dividing two negative values results in a positive value.

$x = - \frac{\ln (793)}{5} + \frac{1}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (793)}{5} + \frac{1}{5}$

Decimal Form:

$x = - 1.13516464 \dots$