Linear Algebra Examples

$\log_{2} (\sqrt[3]{\frac{2 + x}{4 x}})$

Step 1

Set the argument in $\log_{2} (\sqrt[3]{\frac{2 + x}{4 x}})$ greater than $0$ to find where the expression is defined.

$\sqrt[3]{\frac{2 + x}{4 x}} > 0$

Step 2

Solve for $x$ .

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Step 2.1

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

${\sqrt[3]{\frac{2 + x}{4 x}}}^{3} > 0^{3}$

Step 2.2

Simplify each side of the inequality.

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Step 2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{\frac{2 + x}{4 x}}$ as ${(\frac{2 + x}{4 x})}^{\frac{1}{3}}$ .

${({(\frac{2 + x}{4 x})}^{\frac{1}{3}})}^{3} > 0^{3}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Simplify ${({(\frac{2 + x}{4 x})}^{\frac{1}{3}})}^{3}$ .

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Step 2.2.2.1.1

Multiply the exponents in ${({(\frac{2 + x}{4 x})}^{\frac{1}{3}})}^{3}$ .

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Step 2.2.2.1.1.1

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

${(\frac{2 + x}{4 x})}^{\frac{1}{3} \cdot 3} > 0^{3}$

Step 2.2.2.1.1.2

Cancel the common factor of $3$ .

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Step 2.2.2.1.1.2.1

Cancel the common factor.

${(\frac{2 + x}{4 x})}^{\frac{1}{3} \cdot 3} > 0^{3}$

Step 2.2.2.1.1.2.2

Rewrite the expression.

${(\frac{2 + x}{4 x})}^{1} > 0^{3}$

Step 2.2.2.1.2

Simplify.

$\frac{2 + x}{4 x} > 0^{3}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Raising $0$ to any positive power yields $0$ .

$\frac{2 + x}{4 x} > 0$

Step 2.3

Solve for $x$ .

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Step 2.3.1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$4 x = 0$

$2 + x = 0$

Step 2.3.2

Divide each term in $4 x = 0$ by $4$ and simplify.

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Step 2.3.2.1

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Step 2.3.2.2

Simplify the left side.

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Step 2.3.2.2.1

Cancel the common factor of $4$ .

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Step 2.3.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Step 2.3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Step 2.3.2.3

Simplify the right side.

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Step 2.3.2.3.1

Divide $0$ by $4$ .

$x = 0$

Step 2.3.3

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.3.4

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = - 2$

Step 2.3.5

Consolidate the solutions.

$x = 0, - 2$

Step 2.4

Find the domain of $\sqrt[3]{\frac{2 + x}{4 x}}$ .

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Step 2.4.1

Set the denominator in $\frac{2 + x}{4 x}$ equal to $0$ to find where the expression is undefined.

$4 x = 0$

Step 2.4.2

Divide each term in $4 x = 0$ by $4$ and simplify.

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Step 2.4.2.1

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Step 2.4.2.2

Simplify the left side.

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Step 2.4.2.2.1

Cancel the common factor of $4$ .

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Step 2.4.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Step 2.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Step 2.4.2.3

Simplify the right side.

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Step 2.4.2.3.1

Divide $0$ by $4$ .

$x = 0$

Step 2.4.3

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

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

Step 2.5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$x > 0$

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 $x < - 2$ to see if it makes the inequality true.

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Step 2.6.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 2.6.1.2

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

$\sqrt[3]{\frac{2 - 4}{4 (- 4)}} > 0$

Step 2.6.1.3

The left side $0.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.2

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

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Step 2.6.2.1

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 2.6.2.2

Replace $x$ with $- 1$ in the original inequality.

$\sqrt[3]{\frac{2 - 1}{4 (- 1)}} > 0$

Step 2.6.2.3

The left side $- 0.62996052$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 2.6.3

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

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Step 2.6.3.1

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

$x = 2$

Step 2.6.3.2

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

$\sqrt[3]{\frac{2 + 2}{4 (2)}} > 0$

Step 2.6.3.3

The left side $0.79370052$ 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.

$x < - 2$ True

$- 2 < x < 0$ False

$x > 0$ True

$x < - 2$ True

$- 2 < x < 0$ False

$x > 0$ True

Step 2.7

The solution consists of all of the true intervals.

$x < - 2$ or $x > 0$

Step 3

Set the denominator in $\frac{2 + x}{4 x}$ equal to $0$ to find where the expression is undefined.

$4 x = 0$

Step 4

Divide each term in $4 x = 0$ by $4$ and simplify.

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Step 4.1

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $4$ .

$x = 0$

Step 5

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

Interval Notation:

$(- \infty, - 2) \cup (0, \infty)$

Set-Builder Notation:

${x | x < - 2, x > 0}$

Step 6