Precalculus Examples

$\log_{5} (x) + \log_{5} (y) = \log_{5} (3) + 1$

Step 1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{5} (x y) = \log_{5} (3) + 1$

Step 2

Move all the terms containing a logarithm to the left side of the equation.

$\log_{5} (x y) - \log_{5} (3) = 1$

Step 3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{5} (\frac{x y}{3}) = 1$

Step 4

Rewrite $\log_{5} (\frac{x y}{3}) = 1$ 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 $b^{y} = x$ .

$5^{1} = \frac{x y}{3}$

Step 5

Solve for $y$ .

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Step 5.1

Rewrite the equation as $\frac{x y}{3} = 5^{1}$ .

$\frac{x y}{3} = 5$

Step 5.2

Multiply both sides of the equation by $3$ .

$3 \frac{x y}{3} = 3 \cdot 5^{1}$

Step 5.3

Simplify both sides of the equation.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Cancel the common factor of $3$ .

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Step 5.3.1.1.1

Cancel the common factor.

$3 \frac{x y}{3} = 3 \cdot 5^{1}$

Step 5.3.1.1.2

Rewrite the expression.

$x y = 3 \cdot 5^{1}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Simplify $3 \cdot 5^{1}$ .

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Step 5.3.2.1.1

Evaluate the exponent.

$x y = 3 \cdot 5$

Step 5.3.2.1.2

Multiply $3$ by $5$ .

$x y = 15$

Step 5.4

Divide each term in $x y = 15$ by $x$ and simplify.

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Step 5.4.1

Divide each term in $x y = 15$ by $x$ .

$\frac{x y}{x} = \frac{15}{x}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $x$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{x y}{x} = \frac{15}{x}$

Step 5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{15}{x}$