Precalculus Examples

$y > 4^{x}$ , $y < 8 - x^{2}$ , $x + y > 0$

Step 1

Simplify the first inequality.

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Step 1.1

Rewrite so $x$ is on the left side of the inequality.

$4^{x} < y$ and $y < 8 - x^{2}$

Step 1.2

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

$\ln (4^{x}) < \ln (y)$ and $y < 8 - x^{2}$

Step 1.3

Expand $\ln (4^{x})$ by moving $x$ outside the logarithm.

$x \ln (4) < \ln (y)$ and $y < 8 - x^{2}$

Step 1.4

Divide each term in $x \ln (4) < \ln (y)$ by $\ln (4)$ and simplify.

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Step 1.4.1

Divide each term in $x \ln (4) < \ln (y)$ by $\ln (4)$ .

$\frac{x \ln (4)}{\ln (4)} < \frac{\ln (y)}{\ln (4)}$ and $y < 8 - x^{2}$

Step 1.4.2

Simplify the left side.

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Step 1.4.2.1

Cancel the common factor of $\ln (4)$ .

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Step 1.4.2.1.1

Cancel the common factor.

$\frac{x \ln (4)}{\ln (4)} < \frac{\ln (y)}{\ln (4)}$ and $y < 8 - x^{2}$

Step 1.4.2.1.2

Divide $x$ by $1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < 8 - x^{2}$

Step 2

Simplify the second inequality.

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Step 2.1

Rewrite so $x$ is on the left side of the inequality.

$x < \frac{\ln (y)}{\ln (4)}$ and $8 - x^{2} > y$

Step 2.2

Subtract $8$ from both sides of the inequality.

$x < \frac{\ln (y)}{\ln (4)}$ and $- x^{2} > y - 8$

Step 2.3

Divide each term in $- x^{2} > y - 8$ by $- 1$ and simplify.

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Step 2.3.1

Divide each term in $- x^{2} > y - 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$x < \frac{\ln (y)}{\ln (4)}$ and $\frac{- x^{2}}{- 1} < \frac{y}{- 1} + \frac{- 8}{- 1}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Dividing two negative values results in a positive value.

$x < \frac{\ln (y)}{\ln (4)}$ and $\frac{x^{2}}{1} < \frac{y}{- 1} + \frac{- 8}{- 1}$

Step 2.3.2.2

Divide $x^{2}$ by $1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $x^{2} < \frac{y}{- 1} + \frac{- 8}{- 1}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Simplify each term.

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Step 2.3.3.1.1

Move the negative one from the denominator of $\frac{y}{- 1}$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $x^{2} < - 1 \cdot y + \frac{- 8}{- 1}$

Step 2.3.3.1.2

Rewrite $- 1 \cdot y$ as $- y$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $x^{2} < - y + \frac{- 8}{- 1}$

Step 2.3.3.1.3

Divide $- 8$ by $- 1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $x^{2} < - y + 8$

Step 2.4

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$x < \frac{\ln (y)}{\ln (4)}$ and $\sqrt{x^{2}} < \sqrt{- y + 8}$

Step 2.5

Simplify the left side.

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Step 2.5.1

Pull terms out from under the radical.

$x < \frac{\ln (y)}{\ln (4)}$ and $| x | < \sqrt{- y + 8}$

Step 2.6

Write $| x | < \sqrt{- y + 8}$ as a piecewise.

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Step 2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x < \sqrt{- y + 8}$

Step 2.6.3

Find the domain of $x < \sqrt{- y + 8}$ and find the intersection with $x \geq 0$ .

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Step 2.6.3.1

Find the domain of $x < \sqrt{- y + 8}$ .

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Step 2.6.3.1.1

Set the radicand in $\sqrt{- y + 8}$ greater than or equal to $0$ to find where the expression is defined.

$- y + 8 \geq 0$

Step 2.6.3.1.2

Solve for $x$ .

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Step 2.6.3.1.2.1

Subtract $8$ from both sides of the inequality.

$- y \geq - 8$

Step 2.6.3.1.2.2

Divide each term in $- y \geq - 8$ by $- 1$ and simplify.

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Step 2.6.3.1.2.2.1

Divide each term in $- y \geq - 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \leq \frac{- 8}{- 1}$

Step 2.6.3.1.2.2.2

Simplify the left side.

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Step 2.6.3.1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{- 8}{- 1}$

Step 2.6.3.1.2.2.2.2

Divide $y$ by $1$ .

$y \leq \frac{- 8}{- 1}$

Step 2.6.3.1.2.2.3

Simplify the right side.

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Step 2.6.3.1.2.2.3.1

Divide $- 8$ by $- 1$ .

$y \leq 8$

Step 2.6.3.1.3

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

$(- \infty, 8]$

Step 2.6.3.2

Find the intersection of $x \geq 0$ and $(- \infty, 8]$ .

$0 \leq x \leq 8$

Step 2.6.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.6.5

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x < \sqrt{- y + 8}$

Step 2.6.6

Find the domain of $- x < \sqrt{- y + 8}$ and find the intersection with $x < 0$ .

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Step 2.6.6.1

Find the domain of $- x < \sqrt{- y + 8}$ .

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Step 2.6.6.1.1

Set the radicand in $\sqrt{- y + 8}$ greater than or equal to $0$ to find where the expression is defined.

$- y + 8 \geq 0$

Step 2.6.6.1.2

Solve for $x$ .

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Step 2.6.6.1.2.1

Subtract $8$ from both sides of the inequality.

$- y \geq - 8$

Step 2.6.6.1.2.2

Divide each term in $- y \geq - 8$ by $- 1$ and simplify.

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Step 2.6.6.1.2.2.1

Divide each term in $- y \geq - 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \leq \frac{- 8}{- 1}$

Step 2.6.6.1.2.2.2

Simplify the left side.

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Step 2.6.6.1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{- 8}{- 1}$

Step 2.6.6.1.2.2.2.2

Divide $y$ by $1$ .

$y \leq \frac{- 8}{- 1}$

Step 2.6.6.1.2.2.3

Simplify the right side.

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Step 2.6.6.1.2.2.3.1

Divide $- 8$ by $- 1$ .

$y \leq 8$

Step 2.6.6.1.3

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

$(- \infty, 8]$

Step 2.6.6.2

Find the intersection of $x < 0$ and $(- \infty, 8]$ .

$x < 0$

Step 2.6.7

Write as a piecewise.

$x < \frac{\ln (y)}{\ln (4)}$ and ${\begin{cases} x < \sqrt{- y + 8} & 0 \leq x \leq 8 \\ - x < \sqrt{- y + 8} & x < 0 \end{cases}$

Step 2.7

Solve $x < \sqrt{- y + 8}$ when $0 \leq x \leq 8$ .

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Step 2.7.1

Solve $x < \sqrt{- y + 8}$ for $y$ .

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Step 2.7.1.1

Rewrite so $y$ is on the left side of the inequality.

$x < \frac{\ln (y)}{\ln (4)}$ and $\sqrt{- y + 8} > x$

Step 2.7.1.2

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

$x < \frac{\ln (y)}{\ln (4)}$ and ${\sqrt{- y + 8}}^{2} > x^{2}$

Step 2.7.1.3

Simplify each side of the inequality.

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Step 2.7.1.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- y + 8}$ as ${(- y + 8)}^{\frac{1}{2}}$ .

$x < \frac{\ln (y)}{\ln (4)}$ and ${({(- y + 8)}^{\frac{1}{2}})}^{2} > x^{2}$

Step 2.7.1.3.2

Simplify the left side.

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Step 2.7.1.3.2.1

Simplify ${({(- y + 8)}^{\frac{1}{2}})}^{2}$ .

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Step 2.7.1.3.2.1.1

Multiply the exponents in ${({(- y + 8)}^{\frac{1}{2}})}^{2}$ .

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Step 2.7.1.3.2.1.1.1

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

$x < \frac{\ln (y)}{\ln (4)}$ and ${(- y + 8)}^{\frac{1}{2} \cdot 2} > x^{2}$

Step 2.7.1.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.7.1.3.2.1.1.2.1

Cancel the common factor.

$x < \frac{\ln (y)}{\ln (4)}$ and ${(- y + 8)}^{\frac{1}{2} \cdot 2} > x^{2}$

Step 2.7.1.3.2.1.1.2.2

Rewrite the expression.

$x < \frac{\ln (y)}{\ln (4)}$ and $- y + 8 > x^{2}$

Step 2.7.1.3.2.1.2

Simplify.

$x < \frac{\ln (y)}{\ln (4)}$ and $- y + 8 > x^{2}$

Step 2.7.1.4

Solve for $y$ .

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Step 2.7.1.4.1

Subtract $8$ from both sides of the inequality.

$x < \frac{\ln (y)}{\ln (4)}$ and $- y > x^{2} - 8$

Step 2.7.1.4.2

Divide each term in $- y > x^{2} - 8$ by $- 1$ and simplify.

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Step 2.7.1.4.2.1

Divide each term in $- y > x^{2} - 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$x < \frac{\ln (y)}{\ln (4)}$ and $\frac{- y}{- 1} < \frac{x^{2}}{- 1} + \frac{- 8}{- 1}$

Step 2.7.1.4.2.2

Simplify the left side.

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Step 2.7.1.4.2.2.1

Dividing two negative values results in a positive value.

$x < \frac{\ln (y)}{\ln (4)}$ and $\frac{y}{1} < \frac{x^{2}}{- 1} + \frac{- 8}{- 1}$

Step 2.7.1.4.2.2.2

Divide $y$ by $1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < \frac{x^{2}}{- 1} + \frac{- 8}{- 1}$

Step 2.7.1.4.2.3

Simplify the right side.

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Step 2.7.1.4.2.3.1

Simplify each term.

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Step 2.7.1.4.2.3.1.1

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < - 1 \cdot x^{2} + \frac{- 8}{- 1}$

Step 2.7.1.4.2.3.1.2

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < - x^{2} + \frac{- 8}{- 1}$

Step 2.7.1.4.2.3.1.3

Divide $- 8$ by $- 1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < - x^{2} + 8$

Step 2.7.2

Find the intersection of $y < - x^{2} + 8$ and $0 \leq x \leq 8$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $0 \leq x \leq N o$ Minimum

Step 2.8

Solve $- x < \sqrt{- y + 8}$ when $x < 0$ .

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Step 2.8.1

Solve $- x < \sqrt{- y + 8}$ for $y$ .

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Step 2.8.1.1

Rewrite so $y$ is on the left side of the inequality.

$x < \frac{\ln (y)}{\ln (4)}$ and $\sqrt{- y + 8} > - x$

Step 2.8.1.2

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

$x < \frac{\ln (y)}{\ln (4)}$ and ${\sqrt{- y + 8}}^{2} > {(- x)}^{2}$

Step 2.8.1.3

Simplify each side of the inequality.

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Step 2.8.1.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- y + 8}$ as ${(- y + 8)}^{\frac{1}{2}}$ .

$x < \frac{\ln (y)}{\ln (4)}$ and ${({(- y + 8)}^{\frac{1}{2}})}^{2} > {(- x)}^{2}$

Step 2.8.1.3.2

Simplify the left side.

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Step 2.8.1.3.2.1

Simplify ${({(- y + 8)}^{\frac{1}{2}})}^{2}$ .

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Step 2.8.1.3.2.1.1

Multiply the exponents in ${({(- y + 8)}^{\frac{1}{2}})}^{2}$ .

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Step 2.8.1.3.2.1.1.1

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

$x < \frac{\ln (y)}{\ln (4)}$ and ${(- y + 8)}^{\frac{1}{2} \cdot 2} > {(- x)}^{2}$

Step 2.8.1.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.8.1.3.2.1.1.2.1

Cancel the common factor.

$x < \frac{\ln (y)}{\ln (4)}$ and ${(- y + 8)}^{\frac{1}{2} \cdot 2} > {(- x)}^{2}$

Step 2.8.1.3.2.1.1.2.2

Rewrite the expression.

$x < \frac{\ln (y)}{\ln (4)}$ and $- y + 8 > {(- x)}^{2}$

Step 2.8.1.3.2.1.2

Simplify.

$x < \frac{\ln (y)}{\ln (4)}$ and $- y + 8 > {(- x)}^{2}$

Step 2.8.1.3.3

Simplify the right side.

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Step 2.8.1.3.3.1

Simplify ${(- x)}^{2}$ .

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Step 2.8.1.3.3.1.1

Apply the product rule to $- x$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $- y + 8 > {(- 1)}^{2} x^{2}$

Step 2.8.1.3.3.1.2

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

$x < \frac{\ln (y)}{\ln (4)}$ and $- y + 8 > 1 x^{2}$

Step 2.8.1.3.3.1.3

Multiply $x^{2}$ by $1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $- y + 8 > x^{2}$

Step 2.8.1.4

Solve for $y$ .

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Step 2.8.1.4.1

Subtract $8$ from both sides of the inequality.

$x < \frac{\ln (y)}{\ln (4)}$ and $- y > x^{2} - 8$

Step 2.8.1.4.2

Divide each term in $- y > x^{2} - 8$ by $- 1$ and simplify.

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Step 2.8.1.4.2.1

Divide each term in $- y > x^{2} - 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$x < \frac{\ln (y)}{\ln (4)}$ and $\frac{- y}{- 1} < \frac{x^{2}}{- 1} + \frac{- 8}{- 1}$

Step 2.8.1.4.2.2

Simplify the left side.

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Step 2.8.1.4.2.2.1

Dividing two negative values results in a positive value.

$x < \frac{\ln (y)}{\ln (4)}$ and $\frac{y}{1} < \frac{x^{2}}{- 1} + \frac{- 8}{- 1}$

Step 2.8.1.4.2.2.2

Divide $y$ by $1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < \frac{x^{2}}{- 1} + \frac{- 8}{- 1}$

Step 2.8.1.4.2.3

Simplify the right side.

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Step 2.8.1.4.2.3.1

Simplify each term.

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Step 2.8.1.4.2.3.1.1

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < - 1 \cdot x^{2} + \frac{- 8}{- 1}$

Step 2.8.1.4.2.3.1.2

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < - x^{2} + \frac{- 8}{- 1}$

Step 2.8.1.4.2.3.1.3

Divide $- 8$ by $- 1$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $y < - x^{2} + 8$

Step 2.8.2

Find the intersection of $y < - x^{2} + 8$ and $x < 0$ .

$x < \frac{\ln (y)}{\ln (4)}$ and $x < N o$ Minimum

Step 2.9

Find the union of the solutions.

$x < \frac{\ln (y)}{\ln (4)}$ and $x < N o$ Maximum

Step 3