Precalculus Examples

$\ln (x^{2} - x) = \ln (6)$

Step 1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} - x = 6$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $6$ from both sides of the equation.

$x^{2} - x - 6 = 0$

Step 2.2

Factor $x^{2} - x - 6$ using the AC method.

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Step 2.2.1

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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 2.2.2

Write the factored form using these integers.

$(x - 3) (x + 2) = 0$

Step 2.3

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

$x - 3 = 0$

$x + 2 = 0$

Step 2.4

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

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Step 2.4.1

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

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

The final solution is all the values that make $(x - 3) (x + 2) = 0$ true.

$x = 3, - 2$