Solve for x log base five of x minus five plus log base five of x plus fifteen equals three

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

Clarify the left side.

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

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

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

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Use the distributive law.

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

Apply the distributive property.

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

Apply the distributive property.

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

Clarify and Add like terms.

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Clarify each term.

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First, multiply $x$ by $x$ .

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

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

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

Multiply $- 5$ by $15$ .

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

Subtract $5 x$ from $15 x$ .

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

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

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

Calculate $x$

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

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

Rewrite the equation as $x^{2} + 10 x - 75 = 125$ .

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

Move $125$ to the left side of the equation by subtracting it from both sides.

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

Subtract $125$ from $- 75$ .

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

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

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

$- 10, 20$

Use the integers to write the factored form.

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

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

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Make the factor equal $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

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

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Set the factor equal to $0$ .

$x + 20 = 0$

Subtract $20$ from both sides of the equation.

$x = - 20$

The solution is the result of $x - 10 = 0$ and $x + 20 = 0$ .

$x = 10, - 20$

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

$x = 10$

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