Hi, I am having trouble understanding this part of question 3 in the 2014 level 2 algebra paper, particularly how to go about solving part (iii). How do I solve this?
Kia ora @Alexx
This is a quite common exponential solving question and involves the use of logs. For the first parts:
i) Just wants you to identify that the 0.8 is because it is a 20% reduction (100% - 20%= 80%)
ii) Initial medication can be found by subsituting in t=0, which results in 250.4
The third part is a bit more tricky as we need that answer from ii) and we also need to not exceed the 300mg. If they already have 250.4 that means they can only have an additional 49.6 (300-250.4=49.6)
They tell us the rate is still the same so we will keep that at 0.8 and we are now worried about the time from this second dose so lets call this “t”
Drug = initial amount x rate ^ time
49.6 = 250.4 x 0.8^t
You then go on to solve this new equation to find a value for t.
49.6/250.4 = 0.8^t
Take the log of both sides:
log(49.6/250.4) = log(0.8^t)
Move exponent to the front
log(49.6/250.4) = t*log(0.8)
log(49.6/250.4)/log(0.8) = t
put in your calculator to get answer for t
Hope this helps
This type of question is quite common, have a go at some of the other NCEA papers and if you have any questions we are happy to help.
To solve question (iii) you have to remember, that amount of drug in the blood stream can’t exceed 300mg. Because the initial dose is 250.4mg, the maximum amount of the drug before administrering a new dose cannot be more than 300 - 250.4 = 49.4mg. You need to find the time after which amount of the drug in the blood stream would drop to 49.4mg.
You either can use the original equation so
49.6 = 224 * 0.8^(t-0.5)
Or because you already know the initial amount of the drug administered from (ii) you can write:
49.6 = 250.4 * 0.8 ^t
Then you solve this logarithmic equation using a method you are most familiar with to find t which will equal 7.25hrs. Which means that after the drug was adminstered it cannot be repeated earlier than 7hrs 15min.