Precalculus Examples

$\log_{9} (3 v^{2} - 7 v) = \log_{9} (8 + 2 v^{2})$

Step 1

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

$3 v^{2} - 7 v = 8 + 2 v^{2}$

Step 2

Solve for $v$ .

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Step 2.1

Move all terms containing $v$ to the left side of the equation.

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Step 2.1.1

Subtract $2 v^{2}$ from both sides of the equation.

$3 v^{2} - 7 v - 2 v^{2} = 8$

Step 2.1.2

Subtract $2 v^{2}$ from $3 v^{2}$ .

$v^{2} - 7 v = 8$

Step 2.2

Subtract $8$ from both sides of the equation.

$v^{2} - 7 v - 8 = 0$

Step 2.3

Factor $v^{2} - 7 v - 8$ using the AC method.

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Step 2.3.1

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 $- 8$ and whose sum is $- 7$ .

$- 8, 1$

Step 2.3.2

Write the factored form using these integers.

$(v - 8) (v + 1) = 0$

Step 2.4

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

$v - 8 = 0$

$v + 1 = 0$

Step 2.5

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

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Step 2.5.1

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

$v - 8 = 0$

Step 2.5.2

Add $8$ to both sides of the equation.

$v = 8$

Step 2.6

Set $v + 1$ equal to $0$ and solve for $v$ .

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Step 2.6.1

Set $v + 1$ equal to $0$ .

$v + 1 = 0$

Step 2.6.2

Subtract $1$ from both sides of the equation.

$v = - 1$

Step 2.7

The final solution is all the values that make $(v - 8) (v + 1) = 0$ true.

$v = 8, - 1$