Precalculus Examples

$160 = {(6.8 \cdot 10^{- 3})}^{- 1} \ln (6.8 \cdot 10^{- 3} \cdot (140 t) + 1)$

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, $b = e$ , $x = 6.8 \cdot 10^{- 3} \cdot (140 t) + 1$ , and $y = 160$ .

$b = e$

$x = 6.8 \cdot 10^{- 3} \cdot (140 t) + 1$

$y = 160$

Step 2

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

$e^{160} = 6.8 \cdot 10^{- 3} \cdot (140 t) + 1$