Precalculus Examples

$\log_{4} (x^{2} - 2 x - 15) - \log_{4} (x + 3) = 2$

Step 1

Simplify the left side.

Tap for more steps...

Step 1.1

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

$\log_{4} (\frac{x^{2} - 2 x - 15}{x + 3}) = 2$

Step 1.2

Factor $x^{2} - 2 x - 15$ using the AC method.

Tap for more steps...

Step 1.2.1

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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Step 1.2.2

Write the factored form using these integers.

$\log_{4} (\frac{(x - 5) (x + 3)}{x + 3}) = 2$

Step 1.3

Cancel the common factor of $x + 3$ .

Tap for more steps...

Step 1.3.1

Cancel the common factor.

$\log_{4} (\frac{(x - 5) (x + 3)}{x + 3}) = 2$

Step 1.3.2

Divide $x - 5$ by $1$ .

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

Step 2

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

$4^{2} = x - 5$

Step 3

Solve for $x$ .

Tap for more steps...

Step 3.1

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

$x - 5 = 4^{2}$

Step 3.2

Raise $4$ to the power of $2$ .

$x - 5 = 16$

Step 3.3

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

Tap for more steps...

Step 3.3.1

Add $5$ to both sides of the equation.

$x = 16 + 5$

Step 3.3.2

Add $16$ and $5$ .

$x = 21$