Precalculus Examples

$\log_{5} (x - 5) + \log_{5} (x + 15) = 3$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{5} ((x - 5) (x + 15)) = 3$

Step 1.2

Expand $(x - 5) (x + 15)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log_{5} (x (x + 15) - 5 (x + 15)) = 3$

Step 1.2.2

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 15 - 5 (x + 15)) = 3$

Step 1.2.3

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 15 - 5 x - 5 \cdot 15) = 3$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $x$ by $x$ .

$\log_{5} (x^{2} + x \cdot 15 - 5 x - 5 \cdot 15) = 3$

Step 1.3.1.2

Move $15$ to the left of $x$ .

$\log_{5} (x^{2} + 15 \cdot x - 5 x - 5 \cdot 15) = 3$

Step 1.3.1.3

Multiply $- 5$ by $15$ .

$\log_{5} (x^{2} + 15 x - 5 x - 75) = 3$

Step 1.3.2

Subtract $5 x$ from $15 x$ .

$\log_{5} (x^{2} + 10 x - 75) = 3$

Step 2

Rewrite $\log_{5} (x^{2} + 10 x - 75) = 3$ 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 $b^{y} = x$ .

$5^{3} = x^{2} + 10 x - 75$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} + 10 x - 75 = 5^{3}$ .

$x^{2} + 10 x - 75 = 5^{3}$

Step 3.2

Raise $5$ to the power of $3$ .

$x^{2} + 10 x - 75 = 125$

Step 3.3

Subtract $125$ from both sides of the equation.

$x^{2} + 10 x - 75 - 125 = 0$

Step 3.4

Subtract $125$ from $- 75$ .

$x^{2} + 10 x - 200 = 0$

Step 3.5

Factor $x^{2} + 10 x - 200$ using the AC method.

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Step 3.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 200$ and whose sum is $10$ .

$- 10, 20$

Step 3.5.2

Write the factored form using these integers.

$(x - 10) (x + 20) = 0$

Step 3.6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 10 = 0$

$x + 20 = 0$

Step 3.7

Set $x - 10$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Step 3.7.2

Add $10$ to both sides of the equation.

$x = 10$

Step 3.8

Set $x + 20$ equal to $0$ and solve for $x$ .

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Step 3.8.1

Set $x + 20$ equal to $0$ .

$x + 20 = 0$

Step 3.8.2

Subtract $20$ from both sides of the equation.

$x = - 20$

Step 3.9

The final solution is all the values that make $(x - 10) (x + 20) = 0$ true.

$x = 10, - 20$

Step 4

Exclude the solutions that do not make $\log_{5} (x - 5) + \log_{5} (x + 15) = 3$ true.

$x = 10$