Answer
$a)$ $-10\ln3$
$b)$ $\frac{-1}{k}\ln2$
$c)$ $\frac{\ln0.4}{\ln0.2}$
Work Step by Step
$a)$
$e^{-0.3t}$ = $27$
$-0.3t$ = $\ln27$
$t$ = $\frac{1}{-0.3}\ln27$
$t$ = $\frac{-10}{3}\ln(3^{3})$
$t$ = $\frac{-30}{3}\ln3$
$t$ = $-10\ln3$
$b)$
$e^{kt}$ = $\frac{1}{2}$
$kt$ = $\ln(\frac{1}{2})$
$t$ = $\frac{1}{k}\ln(\frac{1}{2})$
$t$ = $\frac{1}{k}\ln(2^{-1})$
$t$ = $\frac{-1}{k}\ln2$
$c)$
$e^{(\ln0.2)t}$ = $0.4$
$(\ln0.2)t$ = $\ln0.4$
$t$ = $\frac{\ln0.4}{\ln0.2}$
