Solve for x 1-e^x=0

$1 - e^{x} = 0$

Step 1

Subtract $1$ from both sides of the equation.

$- e^{x} = - 1$

Step 2

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

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

$\frac{- e^{x}}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{e^{x}}{1} = \frac{- 1}{- 1}$

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

$e^{x} = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$e^{x} = 1$

Step 3

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

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

Step 4

Expand the left side.

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

$x \ln (e) = \ln (1)$

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

$x \cdot 1 = \ln (1)$

Multiply $x$ by $1$ .

$x = \ln (1)$

Step 5

The natural logarithm of $1$ is $0$ .

$x = 0$