Precalculus Examples

$\ln (n^{2} + 12) = \ln (- 9 n - 2)$

Step 1

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

$n^{2} + 12 = - 9 n - 2$

Step 2

Solve for $n$ .

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Step 2.1

Add $9 n$ to both sides of the equation.

$n^{2} + 12 + 9 n = - 2$

Step 2.2

Add $2$ to both sides of the equation.

$n^{2} + 12 + 9 n + 2 = 0$

Step 2.3

Add $12$ and $2$ .

$n^{2} + 9 n + 14 = 0$

Step 2.4

Factor $n^{2} + 9 n + 14$ using the AC method.

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Step 2.4.1

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 $14$ and whose sum is $9$ .

$2, 7$

Step 2.4.2

Write the factored form using these integers.

$(n + 2) (n + 7) = 0$

Step 2.5

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

$n + 2 = 0$

$n + 7 = 0$

Step 2.6

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

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Step 2.6.1

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

$n + 2 = 0$

Step 2.6.2

Subtract $2$ from both sides of the equation.

$n = - 2$

Step 2.7

Set $n + 7$ equal to $0$ and solve for $n$ .

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Step 2.7.1

Set $n + 7$ equal to $0$ .

$n + 7 = 0$

Step 2.7.2

Subtract $7$ from both sides of the equation.

$n = - 7$

Step 2.8

The final solution is all the values that make $(n + 2) (n + 7) = 0$ true.

$n = - 2, - 7$