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C4 Edexcel Mock Paper Question 7

I don't understand question 7 at all. I've done part a, but am stuck on b and c! Please can someone explain? Thanks!



Reply 1
you can rewrite (0.98)^t as:
t LOG (0.98), so the equation will be rewritten as:
10tLOG(0.98)
then, when you differentiate that, you will get:
log(0.98)x 10(0.98)^t which is what they are looking for!! when they ask you to show that it satisifies the equation, then it should be in the form of the answer you get for part A which is dm/dt= -KM
if you work out log(0.98)x 10(0.98)^t, then log(0.98) will give you a negative number which is the constant (-k) and the 10(0.98)^t is the M.
Hope this helped. If it doesn't...you can ask further. :smile:
Reply 2
how do you get that result for part b?...
Reply 3
lightsout92
how do you get that result for part b?...


well....its a general rule that
y=a^x = e^(x ln a)
so using the chain rule:
dy/dx = e^(x ln a) * lna = a^x ln a

And in general:
d(a^x)/dx = a^x ln a

so for this....we have 10(0.98)^t
10(0.98)^t = 10 * e^(t ln 0.98)
so differenitiating that will be:
ln(0.98) * e^(t ln 0.98) = ln(0.98) * 10(0.98)^t
ln(0.98)= -0.02020270.....this is a minus number, so this is the value of the constant K (as K is minus)
M=10(0.98)t when differentiating that, you will get:

-0.02020270 * 10(0.98)^t
but....M=10(0.98)^t
so we can replace it with M....so when rewritten we get:
-0.02020270 * M (this is in the same format as DM/DT= -KM)
-K * M
Reply 4
where does the 0.98 and 10 come from

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