## College Algebra (6th Edition)

$t \approx 6.2$ $hours$
$$A = A_{0}e^{kt}$$ Let $A$ represent the percentage of the dosage that remains after $t$ hours; $A_{0}$ is therefore $1$. Since we know that Aspirin's half-life is $12$ hours, we can find the value for $k$ as follows: $$0.5 = \frac{1}{2} = e^{12k}$$ $$\ln \frac{1}{2} = \ln 2^{-1} = -\ln 2 = 12k$$ $$-\frac{\ln 2}{12} = k$$ To find the value of $t$ when $A = 0.7$, then: $$0.7 = e^{-\frac{\ln 2}{12} t} = (e^{\ln 2})^{\frac{-t}{12}} = 2^{\frac{-t}{12}}$$ $$\frac{t}{-12} = \log_{2}0.7$$ $$t = -12\log_{2}0.7 \approx 6.2$$