Precalculus Examples

$f (x) = \sqrt{1 + \ln (x)}$

Step 1

Set the argument in $\ln (x)$ greater than $0$ to find where the expression is defined.

$x > 0$

Step 2

Set the radicand in $\sqrt{1 + \ln (x)}$ greater than or equal to $0$ to find where the expression is defined.

$1 + \ln (x) \geq 0$

Step 3

Solve for $x$ .

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Step 3.1

Convert the inequality to an equality.

$1 + \ln (x) = 0$

Step 3.2

Solve the equation.

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Step 3.2.1

Subtract $1$ from both sides of the equation.

$\ln (x) = - 1$

Step 3.2.2

To solve for $x$ , rewrite the equation using properties of logarithms.

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

Step 3.2.3

Rewrite $\ln (x) = - 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- 1} = x$

Step 3.2.4

Solve for $x$ .

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Step 3.2.4.1

Rewrite the equation as $x = e^{- 1}$ .

$x = e^{- 1}$

Step 3.2.4.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.3

Find the domain of $1 + \ln (x)$ .

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Step 3.3.1

Set the argument in $\ln (x)$ greater than $0$ to find where the expression is defined.

$x > 0$

Step 3.3.2

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

$(0, \infty)$

Step 3.4

The solution consists of all of the true intervals.

$x \geq \frac{1}{e}$

Step 4

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

Interval Notation:

$[\frac{1}{e}, \infty)$

Set-Builder Notation:

${x | x \geq \frac{1}{e}}$

Step 5