Precalculus Examples

$\ln (x^{2} - 12 x + 33) = 0$

Step 1

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x^{2} - 12 x + 33)} = e^{0}$

Step 2

Rewrite $\ln (x^{2} - 12 x + 33) = 0$ 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 $b^{y} = x$ .

$e^{0} = x^{2} - 12 x + 33$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} - 12 x + 33 = e^{0}$ .

$x^{2} - 12 x + 33 = e^{0}$

Step 3.2

Anything raised to $0$ is $1$ .

$x^{2} - 12 x + 33 = 1$

Step 3.3

Subtract $1$ from both sides of the equation.

$x^{2} - 12 x + 33 - 1 = 0$

Step 3.4

Subtract $1$ from $33$ .

$x^{2} - 12 x + 32 = 0$

Step 3.5

Factor $x^{2} - 12 x + 32$ using the AC method.

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Step 3.5.1

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 $- 12$ .

$- 8, - 4$

Step 3.5.2

Write the factored form using these integers.

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

Step 3.6

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 3.7

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

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Step 3.7.1

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

$x - 8 = 0$

Step 3.7.2

Add $8$ to both sides of the equation.

$x = 8$

Step 3.8

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

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Step 3.8.1

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

$x - 4 = 0$

Step 3.8.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.9

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

$x = 8, 4$