Knowing the correct formula, in this case:
\(\displaystyle A(t) \ = \ A_0e^{kt}, \ A(0) \ = \ 40 \ = \ A_0e^{k(0)} \ = \ A_0, \ A \ = \ Amount\)
\(\displaystyle Hence, \ A(t) \ = \ 40e^{kt}, \ A(2) \ = \ 20 \ = \ 40e^{2k}\)
\(\displaystyle e^{2k} \ = \ \frac{1}{2}, \ 2k \ = \ -ln(2), \ k \ = \ -\frac{ln(2)}{2}\)
\(\displaystyle Ergo, \ A(t) \ = \ (40)e^{ln(2)^{\frac{-t}{2}}} \ = \ (40)(2^{\frac{-t}{2}})\)
\(\displaystyle Therefore: \ A(3) \ = \ (40)(2^{-3/2}) \ = \ 10\sqrt 2 \ = \ about \ 14 \ grams.\)
Check: We have a half life or two years, so if 40 grams decomposes to 20 grams in two years, the remaining 20 grams will decompose to 10 grams in another two years (all together, a period of 4 years) Hence A(4) should equal 10 grams.
Ergo, A(4) = (40)[(2^(-4/2)] = 10 grams.
