Precalculus Examples

$2 \log (x) - \log (5) = \log (x + 10)$

Step 1

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

$2 \log (x) - \log (5) - \log (x + 10) = 0$

Step 2

Simplify the left side.

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Step 2.1

Simplify $2 \log (x) - \log (5) - \log (x + 10)$ .

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Step 2.1.1

Simplify $2 \log (x)$ by moving $2$ inside the logarithm.

$\log (x^{2}) - \log (5) - \log (x + 10) = 0$

Step 2.1.2

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

$\log (\frac{x^{2}}{5}) - \log (x + 10) = 0$

Step 2.1.3

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

$\log (\frac{\frac{x^{2}}{5}}{x + 10}) = 0$

Step 2.1.4

Multiply the numerator by the reciprocal of the denominator.

$\log (\frac{x^{2}}{5} \cdot \frac{1}{x + 10}) = 0$

Step 2.1.5

Multiply $\frac{x^{2}}{5}$ by $\frac{1}{x + 10}$ .

$\log (\frac{x^{2}}{5 (x + 10)}) = 0$

Step 3

Rewrite $\log (\frac{x^{2}}{5 (x + 10)}) = 0$ 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 $b^{y} = x$ .

$10^{0} = \frac{x^{2}}{5 (x + 10)}$

Step 4

Cross multiply to remove the fraction.

$x^{2} = 10^{0} (5 (x + 10))$

Step 5

Simplify $10^{0} (5 (x + 10))$ .

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Step 5.1

Simplify the expression.

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Step 5.1.1

Anything raised to $0$ is $1$ .

$x^{2} = 1 (5 (x + 10))$

Step 5.1.2

Multiply $5 (x + 10)$ by $1$ .

$x^{2} = 5 (x + 10)$

Step 5.2

Apply the distributive property.

$x^{2} = 5 x + 5 \cdot 10$

Step 5.3

Multiply $5$ by $10$ .

$x^{2} = 5 x + 50$

Step 6

Subtract $5 x$ from both sides of the equation.

$x^{2} - 5 x = 50$

Step 7

Factor $x$ out of $x^{2} - 5 x$ .

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Step 7.1

Factor $x$ out of $x^{2}$ .

$x \cdot x - 5 x = 50$

Step 7.2

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 50$

Step 7.3

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$x (x - 5) = 50$

Step 8

Simplify $x (x - 5)$ .

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Step 8.1

Apply the distributive property.

$x \cdot x + x \cdot - 5 = 50$

Step 8.2

Simplify the expression.

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Step 8.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 5 = 50$

Step 8.2.2

Move $- 5$ to the left of $x$ .

$x^{2} - 5 x = 50$

Step 9

Subtract $50$ from both sides of the equation.

$x^{2} - 5 x - 50 = 0$

Step 10

Factor $x^{2} - 5 x - 50$ using the AC method.

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Step 10.1

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 $- 50$ and whose sum is $- 5$ .

$- 10, 5$

Step 10.2

Write the factored form using these integers.

$(x - 10) (x + 5) = 0$

Step 11

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

$x - 10 = 0$

$x + 5 = 0$

Step 12

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

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Step 12.1

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

$x - 10 = 0$

Step 12.2

Add $10$ to both sides of the equation.

$x = 10$

Step 13

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 13.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 14

The final solution is all the values that make $(x - 10) (x + 5) = 0$ true.

$x = 10, - 5$

Step 15

Exclude the solutions that do not make $2 \log (x) - \log (5) = \log (x + 10)$ true.

$x = 10$