Precalculus Examples

$\log_{12} (v^{2} + 35) = \log_{12} (- 12 v - 1)$

Step 1

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = 12$ , $x = v^{2} + 35$ , and $y = \log_{12} (- 12 v - 1)$ .

$b = 12$

$x = v^{2} + 35$

$y = \log_{12} (- 12 v - 1)$

Step 2

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

$12^{\log_{12} (- 12 v - 1)} = v^{2} + 35$