Precalculus Examples

$\log_{4} (x^{2} + 14 x + 49) - \log_{4} (x + 7)$

Step 1

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

$\log_{4} (\frac{x^{2} + 14 x + 49}{x + 7})$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $49$ as $7^{2}$ .

$\log_{4} (\frac{x^{2} + 14 x + 7^{2}}{x + 7})$

Step 2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$14 x = 2 \cdot x \cdot 7$

Step 2.3

Rewrite the polynomial.

$\log_{4} (\frac{x^{2} + 2 \cdot x \cdot 7 + 7^{2}}{x + 7})$

Step 2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 7$ .

$\log_{4} (\frac{{(x + 7)}^{2}}{x + 7})$

Step 3

Cancel the common factor of ${(x + 7)}^{2}$ and $x + 7$ .

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Step 3.1

Factor $x + 7$ out of ${(x + 7)}^{2}$ .

$\log_{4} (\frac{(x + 7) (x + 7)}{x + 7})$

Step 3.2

Cancel the common factors.

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Step 3.2.1

Multiply by $1$ .

$\log_{4} (\frac{(x + 7) (x + 7)}{(x + 7) \cdot 1})$

Step 3.2.2

Cancel the common factor.

$\log_{4} (\frac{(x + 7) (x + 7)}{(x + 7) \cdot 1})$

Step 3.2.3

Rewrite the expression.

$\log_{4} (\frac{x + 7}{1})$

Step 3.2.4

Divide $x + 7$ by $1$ .

$\log_{4} (x + 7)$

Step 4

Rewrite $\log_{4} (x + 7)$ using the change of base formula.

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Step 4.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 4.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (x + 7)}{\log (4)}$