Precalculus Examples

$\log_{3} (\sqrt{x} + 1) = 1$

Step 1

Set the argument in $\log_{3} (\sqrt{x} + 1)$ greater than $0$ to find where the expression is defined.

$\sqrt{x} + 1 > 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $1$ from both sides of the inequality.

$\sqrt{x} > - 1$

Step 2.2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{x}}^{2} > {(- 1)}^{2}$

Step 2.3

Simplify each side of the inequality.

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Step 2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(x^{\frac{1}{2}})}^{2} > {(- 1)}^{2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} > {(- 1)}^{2}$

Step 2.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.3.2.1.1.2.1

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} > {(- 1)}^{2}$

Step 2.3.2.1.1.2.2

Rewrite the expression.

$x^{1} > {(- 1)}^{2}$

Step 2.3.2.1.2

Simplify.

$x > {(- 1)}^{2}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Raise $- 1$ to the power of $2$ .

$x > 1$

Step 2.4

Find the domain of $\sqrt{x} + 1$ .

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Step 2.4.1

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

$x \geq 0$

Step 2.4.2

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

$[0, \infty)$

Step 2.5

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

Step 2.6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.6.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 2.6.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 2.6.1.2

Replace $x$ with $- 2$ in the original inequality.

$\sqrt{- 2} + 1 > 0$

Step 2.6.1.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 2.6.2

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 2.6.2.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 2.6.2.2

Replace $x$ with $0.5$ in the original inequality.

$\sqrt{0.5} + 1 > 0$

Step 2.6.2.3

The left side $1.70710678$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 2.6.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 2.6.3.2

Replace $x$ with $4$ in the original inequality.

$\sqrt{4} + 1 > 0$

Step 2.6.3.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < 1$ True

$x > 1$ True

$x < 0$ False

$0 < x < 1$ True

$x > 1$ True

Step 2.7

The solution consists of all of the true intervals.

$0 < x < 1$ or $x > 1$

Step 3

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

$x \geq 0$

Step 4

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

Interval Notation:

$(0, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x > 0, x \neq 1}$

Step 5