help maths A.S ,G.S

19a)
Sub x(t) = 50, Am=100 and t=2
50 = 100 x 2^(-2k)
2^-1 = 2^(-2k)
-2k = -1
k= 1/2

19bi)
We should list out to see the GS pattern.
Amount of drug remaining in the blood stream of Leo immediately before the 1st injection = 0ug
Amount of drug remaining in the blood stream of Leo immediately before the 2nd
injection = 100 x 2^(-1/2 x 4) ug
Amount of drug remaining in the blood stream of Leo immediately before the 3rd
injection = [100 x 2^(-1/2 x 4) + 100] x 2^(-1/2 x 4)
= 100 x [2^(-1/2 x 4)]^2 + 100 x 2^(-1/2 x 4)ug
We can deduce that Amount of drug remaining in the blood stream of Leo immediately before the nth injection
= 100 x [2^(-1/2 x 4)]^(n-1) + 100 x [2^(-1/2 x 4)]^(n-2) + ....... + 100 x 2^(-1/2 x 4)
= 100 x 2^(-1/2 x 4) {1 - [2^(-1/2 x 4)]^(n-1)} / [1- 2^(-1/2 x 4)]
=(100/3)[1-(1/4)^(n-1)] ug

19bii
By using bi,
(100/3)[1-(1/4)^(n-1)] > 32
[1-(1/4)^(n-1)] > (24/25)
(1/4)^(n-1) < (1/25)
(n-1)log(1/4)< log (1/25)
(n-1)>2.32
n > 3.32
So the least required injection number = 4

2015-03-07 00:08:51 補充：
被題目FAKE了 . BII應是3
因為第四針前, 佢已經過了32
故此只需三針