Precalculus Examples

$\log_{3} (x^{2} - 3 x - 1) = 3$

Rewrite $\log_{3} (x^{2} - 3 x - 1) = 3$ 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 $b^{y} = x$ .

$3^{3} = x^{2} - 3 x - 1$

Solve for $x$ .

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Rewrite the equation as $x^{2} - 3 x - 1 = 3^{3}$ .

$x^{2} - 3 x - 1 = 3^{3}$

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

$x^{2} - 3 x - 1 = 27$

Subtract $27$ from both sides of the equation.

$x^{2} - 3 x - 1 - 27 = 0$

Subtract $27$ from $- 1$ .

$x^{2} - 3 x - 28 = 0$

Factor $x^{2} - 3 x - 28$ using the AC method.

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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 $- 28$ and whose sum is $- 3$ .

$- 7, 4$

Write the factored form using these integers.

$(x - 7) (x + 4) = 0$

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

$x - 7 = 0$

$x + 4 = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

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Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 7) (x + 4) = 0$ true.

$x = 7, - 4$