Precalculus Examples

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

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

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)

$\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

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

Step 1.2

Expand

(x - 5) (x + 15)

$(x - 5) (x + 15)$ 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

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

Step 1.2.2

Apply the distributive property.

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

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

Step 1.2.3

Apply the distributive property.

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

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

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

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

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

$x$ by

x

$x$ .

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

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

Step 1.3.1.2

Move

15

$15$ to the left of

x

$x$ .

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

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

Step 1.3.1.3

Multiply

- 5

$- 5$ by

15

$15$ .

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

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

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

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

Step 1.3.2

Subtract

5 x

$5 x$ from

15 x

$15 x$ .

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

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

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

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

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

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

Step 2

Rewrite

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

$\log_{5} (x^{2} + 10 x - 75) = 3$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

\log_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

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

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

Step 3

Solve for

x

$x$ .

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Step 3.1

Rewrite the equation as

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

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

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

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

Step 3.2

Raise

5

$5$ to the power of

3

$3$ .

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

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

Step 3.3

Subtract

125

$125$ from both sides of the equation.

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

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

Step 3.4

Subtract

125

$125$ from

- 75

$- 75$ .

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

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

Step 3.5

Factor

x^{2} + 10 x - 200

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

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Step 3.5.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 200

$- 200$ and whose sum is

10

$10$ .

- 10, 20

$- 10, 20$

Step 3.5.2

Write the factored form using these integers.

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

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

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

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

Step 3.6

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 10 = 0

$x - 10 = 0$

x + 20 = 0

$x + 20 = 0$

Step 3.7

Set

x - 10

$x - 10$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.7.1

Set

x - 10

$x - 10$ equal to

0

$0$ .

x - 10 = 0

$x - 10 = 0$

Step 3.7.2

Add

10

$10$ to both sides of the equation.

x = 10

$x = 10$

x = 10

$x = 10$

Step 3.8

Set

x + 20

$x + 20$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.8.1

Set

x + 20

$x + 20$ equal to

0

$0$ .

x + 20 = 0

$x + 20 = 0$

Step 3.8.2

Subtract

20

$20$ from both sides of the equation.

x = - 20

$x = - 20$

x = - 20

$x = - 20$

Step 3.9

The final solution is all the values that make

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

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

x = 10, - 20

$x = 10, - 20$

x = 10, - 20

$x = 10, - 20$

Step 4

Exclude the solutions that do not make

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

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

x = 10

$x = 10$