#### Answer

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.

#### Work Step by Step

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.