Precalculus Examples

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

Step 1

Set the argument in $\ln (1 - e^{x})$ greater than $0$ to find where the expression is defined.

$1 - e^{x} > 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $1$ from both sides of the inequality.

$- e^{x} > - 1$

Step 2.2

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

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Step 2.2.1

Divide each term in $- e^{x} > - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.2.2.2

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

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 1$ by $- 1$ .

$e^{x} < 1$

Step 2.3

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

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

Step 2.4

Expand the left side.

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $x$ by $1$ .

$x < \ln (1)$

Step 2.5

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

$x < 0$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 0)$

Set-Builder Notation:

${x | x < 0}$

Step 4