Precalculus Examples

$\log_{2} (x + 14) - \log_{2} (x - 3) = 1$

Step 1

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

$\log_{2} (\frac{x + 14}{x - 3}) = 1$

Step 2

Rewrite $\log_{2} (\frac{x + 14}{x - 3}) = 1$ 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 $b^{y} = x$ .

$2^{1} = \frac{x + 14}{x - 3}$

Step 3

Cross multiply to remove the fraction.

$x + 14 = 2 (x - 3)$

Step 4

Simplify $2 (x - 3)$ .

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Step 4.1

Apply the distributive property.

$x + 14 = 2 x + 2 \cdot - 3$

Step 4.2

Multiply $2$ by $- 3$ .

$x + 14 = 2 x - 6$

Step 5

Move all terms containing $x$ to the left side of the equation.

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Step 5.1

Subtract $2 x$ from both sides of the equation.

$x + 14 - 2 x = - 6$

Step 5.2

Subtract $2 x$ from $x$ .

$- x + 14 = - 6$

Step 6

Move all terms not containing $x$ to the right side of the equation.

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Step 6.1

Subtract $14$ from both sides of the equation.

$- x = - 6 - 14$

Step 6.2

Subtract $14$ from $- 6$ .

$- x = - 20$

Step 7

Divide each term in $- x = - 20$ by $- 1$ and simplify.

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Step 7.1

Divide each term in $- x = - 20$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 20}{- 1}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 20}{- 1}$

Step 7.2.2

Divide $x$ by $1$ .

$x = \frac{- 20}{- 1}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide $- 20$ by $- 1$ .

$x = 20$