Precalculus Examples

$\log_{2} (x^{2} - 4 x - 28) = 2$

Step 1

Rewrite $\log_{2} (x^{2} - 4 x - 28) = 2$ 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 $b^{y} = x$ .

$2^{2} = x^{2} - 4 x - 28$

Step 2

Solve for $x$ .

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Step 2.1

Rewrite the equation as $x^{2} - 4 x - 28 = 2^{2}$ .

$x^{2} - 4 x - 28 = 2^{2}$

Step 2.2

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

$x^{2} - 4 x - 28 = 4$

Step 2.3

Subtract $4$ from both sides of the equation.

$x^{2} - 4 x - 28 - 4 = 0$

Step 2.4

Subtract $4$ from $- 28$ .

$x^{2} - 4 x - 32 = 0$

Step 2.5

Factor $x^{2} - 4 x - 32$ using the AC method.

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Step 2.5.1

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 $- 32$ and whose sum is $- 4$ .

$- 8, 4$

Step 2.5.2

Write the factored form using these integers.

$(x - 8) (x + 4) = 0$

Step 2.6

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

$x - 8 = 0$

$x + 4 = 0$

Step 2.7

Set $x - 8$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 2.7.2

Add $8$ to both sides of the equation.

$x = 8$

Step 2.8

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.8.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.8.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.9

The final solution is all the values that make $(x - 8) (x + 4) = 0$ true.

$x = 8, - 4$