Precalculus Examples

$\log (13^{- 7 v}) = \log (2^{x + 6})$

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, $b = 10$ , $x = 13^{- 7 v}$ , and $y = \log (2^{x + 6})$ .

$b = 10$

$x = 13^{- 7 v}$

$y = \log (2^{x + 6})$

Step 2

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

$10^{\log (2^{x + 6})} = 13^{- 7 v}$