Precalculus Examples

$\log (11^{- 8 x}) = \log (8^{- x + 2})$

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, $b = 10$ , $x = 11^{- 8 x}$ , and $y = \log (8^{- x + 2})$ .

$b = 10$

$x = 11^{- 8 x}$

$y = \log (8^{- x + 2})$

Step 2

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

$10^{\log (8^{- x + 2})} = 11^{- 8 x}$