## Calculus: Early Transcendentals 8th Edition

a) After 16 days, there will be $\frac{1}{16}$g remaining. b) $m(t)=2^{-\frac{t}{4}}$ c) $\therefore m^{-1}=-4\frac{\ln m}{\ln2}$ The inverse is the amount of time $t$ in days required for the mass to drop to a certain amount $m$ in grams. d) The mass will be reduced to $0.01$g in approximately $27$ days.
a) After 4 days, $m(4)=1\div2$ After 8 days, $m(8)=m(4)\div2=1\div2^2$ After 12 days, $m(12)=m(8)\div2=1\div2^3$ After 16 days, $m(16)=m(12)\div2=1\div2^4=\frac{1}{16}$ After 16 days, there will be $\frac{1}{16}$g remaining. b) From the above, it can be deduced that $m(t)=1\div2^{\frac{t}{4}}=2^{-\frac{t}{4}}$ c) $m=2^{-\frac{t}{4}}$ $\ln m=\ln (2^{-\frac{t}{4}})=-\frac{t}{4}\ln 2$ $\therefore m^{-1}=t=-4\frac{\ln m}{\ln2}$ The inverse is the amount of time $t$ in days required for the mass to drop to a certain amount $m$ in grams. d) let $m=0.01$ $t=-4\frac{\ln0.01}{\ln2}\approx26.58$ days The mass will be reduced to $0.01$g in approximately $27$ days.