Precalculus Examples

$\log_{13} (p^{2} - 4 p) = \log_{13} (24 - 2 p)$

Step 1

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

$p^{2} - 4 p = 24 - 2 p$

Step 2

Solve for $p$ .

Tap for more steps...

Step 2.1

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

Tap for more steps...

Step 2.1.1

Add $2 p$ to both sides of the equation.

$p^{2} - 4 p + 2 p = 24$

Step 2.1.2

Add $- 4 p$ and $2 p$ .

$p^{2} - 2 p = 24$

Step 2.2

Subtract $24$ from both sides of the equation.

$p^{2} - 2 p - 24 = 0$

Step 2.3

Factor $p^{2} - 2 p - 24$ using the AC method.

Tap for more steps...

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 $- 24$ and whose sum is $- 2$ .

$- 6, 4$

Step 2.3.2

Write the factored form using these integers.

$(p - 6) (p + 4) = 0$

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 $0$ .

$p - 6 = 0$

$p + 4 = 0$

Step 2.5

Set $p - 6$ equal to $0$ and solve for $p$ .

Tap for more steps...

Step 2.5.1

Set $p - 6$ equal to $0$ .

$p - 6 = 0$

Step 2.5.2

Add $6$ to both sides of the equation.

$p = 6$

Step 2.6

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

Tap for more steps...

Step 2.6.1

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

$p + 4 = 0$

Step 2.6.2

Subtract $4$ from both sides of the equation.

$p = - 4$

Step 2.7

The final solution is all the values that make $(p - 6) (p + 4) = 0$ true.

$p = 6, - 4$