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C3 MEI - Need answers from textbook, any help?

Hi,

We've been set some homework from booklets containing questions from the MEI C3 and C4 textbook (we're starting A2 a little early).

However I have not got my textbook yet and need the answers to the questions (full working is required, so I'm not just doing this to quickly do my homework).

If anyone could post the textbooks answers to questions 9 and 10, page 16/17 exercise 2A it'd be a great help,

Cheers
(edited 11 years ago)
Reply 1
Avatar for SecondHand
SecondHand
Post the questions and I will help.
Reply 2
Avatar for aznkid66
aznkid66
1
Yeah, couldn't we just discuss answers until we reach a consensus?
Hi,

Can anyone please help me with this:

Ex 4D, Q10 (Pg 88 of the C3/C4 MEI text book)

A sample of radioacitve substance has a mass m at time t, where m=ae^(-bt), (where a and b are positive constants), in appropriate units.

(i) Explain why a graph of ln(m) against t will be a straight line.

(ii) The graph of ln(m) against t (a straight line with negative gradient) passes through (0,3) and (5,1). Find the values of a and b.


I managed to do part (i), but am really stuck on the second part.

I would really appreciate some help with the second part.
Reply 4
Avatar for MartyO
MartyO
7
Original post by Emmessy Squire
Hi,

Can anyone please help me with this:

(ii) The graph of ln(m) against t (a straight line with negative gradient) passes through (0,3) and (5,1). Find the values of a and b.




Using properties of logarithms, you can rewrite ln⁡a⋅e−bt\ln{a\cdot e^{-bt}} as ln⁡a+ln⁡e−bt\ln {a} + \ln {e^{-bt}}. From there, with the second term, you're asking yourself: "What exponent do I put on ee such that ee raised to that power equals e−bte^{-bt}." So,
Unparseable latex formula:

\ln {a} + \ln {e^{-bt}} =\ln {a} -bt}

.

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