Finite Math Examples

$\log_{5} (x - 1) + \log_{5} (x + 119) = 4$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{5} ((x - 1) (x + 119)) = 4$

Step 1.2

Expand $(x - 1) (x + 119)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log_{5} (x (x + 119) - 1 (x + 119)) = 4$

Step 1.2.2

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 119 - 1 (x + 119)) = 4$

Step 1.2.3

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 119 - 1 x - 1 \cdot 119) = 4$

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 $x$ .

$\log_{5} (x^{2} + x \cdot 119 - 1 x - 1 \cdot 119) = 4$

Step 1.3.1.2

Move $119$ to the left of $x$ .

$\log_{5} (x^{2} + 119 \cdot x - 1 x - 1 \cdot 119) = 4$

Step 1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$\log_{5} (x^{2} + 119 x - x - 1 \cdot 119) = 4$

Step 1.3.1.4

Multiply $- 1$ by $119$ .

$\log_{5} (x^{2} + 119 x - x - 119) = 4$

Step 1.3.2

Subtract $x$ from $119 x$ .

$\log_{5} (x^{2} + 118 x - 119) = 4$

Step 2

Rewrite $\log_{5} (x^{2} + 118 x - 119) = 4$ 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 $b^{y} = x$ .

$5^{4} = x^{2} + 118 x - 119$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} + 118 x - 119 = 5^{4}$ .

$x^{2} + 118 x - 119 = 5^{4}$

Step 3.2

Raise $5$ to the power of $4$ .

$x^{2} + 118 x - 119 = 625$

Step 3.3

Subtract $625$ from both sides of the equation.

$x^{2} + 118 x - 119 - 625 = 0$

Step 3.4

Subtract $625$ from $- 119$ .

$x^{2} + 118 x - 744 = 0$

Step 3.5

Factor $x^{2} + 118 x - 744$ 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 $- 744$ and whose sum is $118$ .

$- 6, 124$

Step 3.5.2

Write the factored form using these integers.

$(x - 6) (x + 124) = 0$

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 $0$ .

$x - 6 = 0$

$x + 124 = 0$

Step 3.7

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 3.7.2

Add $6$ to both sides of the equation.

$x = 6$

Step 3.8

Set $x + 124$ equal to $0$ and solve for $x$ .

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Step 3.8.1

Set $x + 124$ equal to $0$ .

$x + 124 = 0$

Step 3.8.2

Subtract $124$ from both sides of the equation.

$x = - 124$

Step 3.9

The final solution is all the values that make $(x - 6) (x + 124) = 0$ true.

$x = 6, - 124$

Step 4

Exclude the solutions that do not make $\log_{5} (x - 1) + \log_{5} (x + 119) = 4$ true.

$x = 6$