Precalculus Examples

$\ln (0.29) = \ln (e^{- 0.000121 t})$

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, $b = e$ , $x = 0.29$ , and $y = \ln (e^{- 0.000121 t})$ .

$b = e$

$x = 0.29$

$y = \ln (e^{- 0.000121 t})$

Step 2

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

$e^{\ln (e^{- 0.000121 t})} = 0.29$