Precalculus Examples

$\log_{12} (r^{2} + 1) = \log_{12} (- 6 r + 1)$

Step 1

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

$r^{2} + 1 = - 6 r + 1$

Step 2

Solve for $r$ .

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Step 2.1

Add $6 r$ to both sides of the equation.

$r^{2} + 1 + 6 r = 1$

Step 2.2

Subtract $1$ from both sides of the equation.

$r^{2} + 1 + 6 r - 1 = 0$

Step 2.3

Combine the opposite terms in $r^{2} + 1 + 6 r - 1$ .

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Step 2.3.1

Subtract $1$ from $1$ .

$r^{2} + 6 r + 0 = 0$

Step 2.3.2

Add $r^{2} + 6 r$ and $0$ .

$r^{2} + 6 r = 0$

Step 2.4

Factor $r$ out of $r^{2} + 6 r$ .

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Step 2.4.1

Factor $r$ out of $r^{2}$ .

$r \cdot r + 6 r = 0$

Step 2.4.2

Factor $r$ out of $6 r$ .

$r \cdot r + r \cdot 6 = 0$

Step 2.4.3

Factor $r$ out of $r \cdot r + r \cdot 6$ .

$r (r + 6) = 0$

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 $0$ .

$r = 0$

$r + 6 = 0$

Step 2.6

Set $r$ equal to $0$ .

$r = 0$

Step 2.7

Set $r + 6$ equal to $0$ and solve for $r$ .

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Step 2.7.1

Set $r + 6$ equal to $0$ .

$r + 6 = 0$

Step 2.7.2

Subtract $6$ from both sides of the equation.

$r = - 6$

Step 2.8

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

$r = 0, - 6$