Precalculus Examples

$\log (x^{2} + 2 x - 2) = 0$

Step 1

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

$10^{0} = x^{2} + 2 x - 2$

Step 2

Solve for $x$ .

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Step 2.1

Rewrite the equation as $x^{2} + 2 x - 2 = 10^{0}$ .

$x^{2} + 2 x - 2 = 10^{0}$

Step 2.2

Anything raised to $0$ is $1$ .

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

Step 2.3

Subtract $1$ from both sides of the equation.

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

Step 2.4

Subtract $1$ from $- 2$ .

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

Step 2.5

Factor $x^{2} + 2 x - 3$ 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 $- 3$ and whose sum is $2$ .

$- 1, 3$

Step 2.5.2

Write the factored form using these integers.

$(x - 1) (x + 3) = 0$

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 $0$ .

$x - 1 = 0$

$x + 3 = 0$

Step 2.7

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.7.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.8

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.8.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.8.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.9

The final solution is all the values that make $(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$