Precalculus Examples

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

Step 1

Solve for $\sqrt{\ln (x + y)}$ .

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Step 1.1

Rewrite the equation as $\sqrt{\ln (x + y)} = f (x, y)$ .

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

Step 1.2

Multiply $f$ by each element of the matrix.

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

Step 2

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

${\sqrt{\ln (x + y)}}^{2} = {(f x, f y)}^{2}$

Step 3

Simplify each side of the equation.

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Step 3.1

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

${({\ln (x + y)}^{\frac{1}{2}})}^{2} = {(f x, f y)}^{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

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

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Step 3.2.1.1

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

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Step 3.2.1.1.1

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

${\ln (x + y)}^{\frac{1}{2} \cdot 2} = {(f x, f y)}^{2}$

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

${\ln (x + y)}^{\frac{1}{2} \cdot 2} = {(f x, f y)}^{2}$

Step 3.2.1.1.2.2

Rewrite the expression.

$\ln^{1} (x + y) = {(f x, f y)}^{2}$

Step 3.2.1.2

Simplify.

$\ln (x + y) = {(f x, f y)}^{2}$

Step 4

Solve for $y$ .

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Step 4.1

Subtract ${(f x, f y)}^{2}$ from both sides of the equation.

$\ln (x + y) - {(f x, f y)}^{2} = 0$

Step 4.2

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

$e^{\ln (x + y)} = e^{{(f x, f y)}^{2}}$

Step 4.3

Rewrite $\ln (x + y) = {(f x, f y)}^{2}$ 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 $b^{y} = x$ .

$e^{{(f x, f y)}^{2}} = x + y$

Step 4.4

Solve for $y$ .

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Step 4.4.1

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

$\ln (e^{{(f x, f y)}^{2}}) = \ln (x + y)$

Step 4.4.2

Expand the left side.

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Step 4.4.2.1

Expand $\ln (e^{{(f x, f y)}^{2}})$ by moving ${(f x, f y)}^{2}$ outside the logarithm.

${(f x, f y)}^{2} \ln (e) = \ln (x + y)$

Step 4.4.2.2

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

${(f x, f y)}^{2} \cdot 1 = \ln (x + y)$

Step 4.4.2.3

Multiply ${(f x, f y)}^{2}$ by $1$ .

${(f x, f y)}^{2} = \ln (x + y)$

Step 4.4.3

Subtract $\ln (x + y)$ from both sides of the equation.

${(f x, f y)}^{2} - \ln (x + y) = 0$

Step 4.4.4

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

$e^{\ln (x + y)} = e^{{(f x, f y)}^{2}}$

Step 4.4.5

$e^{{(f x, f y)}^{2}} = x + y$

Step 4.4.6

Solve for $y$ .

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Step 4.4.6.1

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

$\ln (e^{{(f x, f y)}^{2}}) = \ln (x + y)$

Step 4.4.6.2

Expand the left side.

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Step 4.4.6.2.1

Expand $\ln (e^{{(f x, f y)}^{2}})$ by moving ${(f x, f y)}^{2}$ outside the logarithm.

${(f x, f y)}^{2} \ln (e) = \ln (x + y)$

Step 4.4.6.2.2

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

${(f x, f y)}^{2} \cdot 1 = \ln (x + y)$

Step 4.4.6.2.3

Multiply ${(f x, f y)}^{2}$ by $1$ .

${(f x, f y)}^{2} = \ln (x + y)$

Step 4.4.6.3

Subtract $\ln (x + y)$ from both sides of the equation.

${(f x, f y)}^{2} - \ln (x + y) = 0$

Step 4.4.6.4

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

$e^{\ln (x + y)} = e^{{(f x, f y)}^{2}}$

Step 4.4.6.5

$e^{{(f x, f y)}^{2}} = x + y$

Step 4.4.6.6

Solve for $y$ .

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Step 4.4.6.6.1

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

$\ln (e^{{(f x, f y)}^{2}}) = \ln (x + y)$

Step 4.4.6.6.2

Expand the left side.

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Step 4.4.6.6.2.1

Expand $\ln (e^{{(f x, f y)}^{2}})$ by moving ${(f x, f y)}^{2}$ outside the logarithm.

${(f x, f y)}^{2} \ln (e) = \ln (x + y)$

Step 4.4.6.6.2.2

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

${(f x, f y)}^{2} \cdot 1 = \ln (x + y)$

Step 4.4.6.6.2.3

Multiply ${(f x, f y)}^{2}$ by $1$ .

${(f x, f y)}^{2} = \ln (x + y)$

Step 4.4.6.6.3

Subtract $\ln (x + y)$ from both sides of the equation.

${(f x, f y)}^{2} - \ln (x + y) = 0$

Step 4.4.6.6.4

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

$e^{\ln (x + y)} = e^{{(f x, f y)}^{2}}$

Step 4.4.6.6.5

$e^{{(f x, f y)}^{2}} = x + y$

Step 4.4.6.6.6

Solve for $y$ .

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Step 4.4.6.6.6.1

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

$\ln (e^{{(f x, f y)}^{2}}) = \ln (x + y)$

Step 4.4.6.6.6.2

Expand the left side.

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Step 4.4.6.6.6.2.1

Expand $\ln (e^{{(f x, f y)}^{2}})$ by moving ${(f x, f y)}^{2}$ outside the logarithm.

${(f x, f y)}^{2} \ln (e) = \ln (x + y)$

Step 4.4.6.6.6.2.2

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

${(f x, f y)}^{2} \cdot 1 = \ln (x + y)$

Step 4.4.6.6.6.2.3

Multiply ${(f x, f y)}^{2}$ by $1$ .

${(f x, f y)}^{2} = \ln (x + y)$

Step 5

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

$x + y > 0$

Step 6

Subtract $y$ from both sides of the inequality.

$x > - y$

Step 7

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$