Answer
12.5 days
Work Step by Step
Exponential growth and decay models are given by $A=A_{0}e^{kt}$
in which $t$ represents time,
$A_{0}$ is the amount present at $t=0$, and
$A$ is the amount present at time $t$.
If $k>0$, the model describes growth and $k$ is the growth rate.
If $k<0$, the model describes decay and $k$ is the decay rate.
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Half-life is the time needed for the substance to decay to $\displaystyle \frac{A_{0}}{2}.$
Use this to find k:
0.5$A_{0}=A_{0}e^{k\cdot 7.2}\qquad/\div A_{0}$
$0.5=e^{7.2k}\qquad$... apply ln( ) to both sides...
$\ln 0.5=7.2k\qquad /\div 7.2$
$ k=\displaystyle \frac{\ln 0.5}{7.2}\approx$-0.0962704417444$\approx-0.09627$
Now, for the time needed for $A=0.30A_{0}, $solve for t:
$0.30A_{0}=A_{0}e^{-0.09627\cdot t}\qquad/\div A_{0}$
$ 0.3=e^{-0.09627\cdot t}\qquad$... apply ln( ) to both sides...
$\ln 0.3=-0.09627\cdot t \qquad/\div(-0.09627)$
$ t=\displaystyle \frac{\ln 0.3}{-0.09627}\approx$12.5062096637$\approx 12.5$ (days)
