Precalculus Examples

$\log_{2} (x^{2} - 4 x - 37) = 3$

Step 1

Rewrite $\log_{2} (x^{2} - 4 x - 37) = 3$ 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 $b^{y} = x$ .

$2^{3} = x^{2} - 4 x - 37$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Rewrite the equation as $x^{2} - 4 x - 37 = 2^{3}$ .

$x^{2} - 4 x - 37 = 2^{3}$

Step 2.2

Raise $2$ to the power of $3$ .

$x^{2} - 4 x - 37 = 8$

Step 2.3

Subtract $8$ from both sides of the equation.

$x^{2} - 4 x - 37 - 8 = 0$

Step 2.4

Subtract $8$ from $- 37$ .

$x^{2} - 4 x - 45 = 0$

Step 2.5

Factor $x^{2} - 4 x - 45$ using the AC method.

Tap for more steps...

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 $- 45$ and whose sum is $- 4$ .

$- 9, 5$

Step 2.5.2

Write the factored form using these integers.

$(x - 9) (x + 5) = 0$

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 $0$ .

$x - 9 = 0$

$x + 5 = 0$

Step 2.7

Set $x - 9$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.7.1

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Step 2.7.2

Add $9$ to both sides of the equation.

$x = 9$

Step 2.8

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

Tap for more steps...

Step 2.8.1

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

$x + 5 = 0$

Step 2.8.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.9

The final solution is all the values that make $(x - 9) (x + 5) = 0$ true.

$x = 9, - 5$