## University Calculus: Early Transcendentals (3rd Edition)

After $0.585$ days, or over half a day, the radon would fall to $90\%$ of the original value.
The decay equation for radon-222 gas is $$y=y_0e^{-0.18t}$$ For the radon to fall to $90\%$ of the original value, it means $y=90\%y_0=0.9y_0$ To find $t$, substitute $y=0.9y_0$ back to the equation: $$0.9y_0=y_0e^{-0.18t}$$ $$e^{-0.18t}=0.9$$ Take the natural logarithm of both sides: $$-0.18t=\ln0.9$$ $$t=-\frac{\ln0.9}{0.18}\approx0.585(days)$$ So after $0.585$ days, or over half a day, the radon would fall to $90\%$ of the original value.