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I've just been accepted to take my PHD in chemical engineering in Melbourne next year. Some how I have gone from the age of 17 with out taking too many extra maths classes and so at the moment (I'm 26) I am trying to teach myself some logarithms and exponentials. Everything is going well so far, I am understanding the laws and piecing them together. Every now and then I come across a question which throws me, one where I am unsure of my methodology, but I'm sure this is where the learning happens.

For example, I just found this question in a textbook:

The mass xkg of a radio-active substance remaining in a sample t days after starting timing is given by the equation x = 4e^-0.2t.

If it asks me to find the mass at the start of the timing, I'm quite certain I just remove the -0.2t exponential right?

Would x = 4ekg be a sufficient answer?

New Zealand's finest
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2 Answers2

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No, the mass is $4e^{-0.2t}$ When you substitute in $t=0$ to get the beginning mass, this becomes $4e^0$. As $e^0=1$, this is $4$, not $4e$

Ross Millikan
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  • Thank you for your help with this. What kind of method could I use in order to find the mass left after 5 years, if mass left t days after starting timing is not specified? – New Zealand's finest May 12 '16 at 01:25
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    What information do you have in that case? Without such an equation as you gave in the original question, you have no information to know the mass left – Ross Millikan May 12 '16 at 02:28
  • Following on from the first question I am just asked to 'find the mass left after 5 years' – New Zealand's finest May 12 '16 at 03:26
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    In this case the timing is the same, so just putting 5 years in the time variable with the appropriate units ($5*(number of days in a year)$) will give you the answer, the thing is, for any time that function gives you the mass at that specific time $(t)$ – Ivan Lerner May 12 '16 at 03:29
  • Great thanks this is very helpful. So just 4e^5 kg *365 – New Zealand's finest May 12 '16 at 04:40
  • Please be more careful. As I commented on your other question, this should read as $4e^5 \cdot 365 \approx 216683$ kg. You lost putting the $365$ in the exponent, changed $0.2$ to $5$, as well as the minus sign in the exponent. It should be $4e^{(-0.2 \cdot 365)}\approx 8\cdot 10^{-32}$ – Ross Millikan May 12 '16 at 05:41
  • But how could it be the latter, when -0.2 is referred to as "t days" and not 5? I'm really sorry if I'm annoying you, but I'm annoying myself because I don't understand.

    I can see that the answer is 216683kg, but I can't see the parity between this and 8.10^-32 kg

     – New Zealand's finest May 13 '16 at 09:27
  • The exponent needs the factor $-0.2$. After five years, it should be $e^{-0.2 \cdot 5 \cdot 365}$ I lost the $5$. The $t$ is in the exponent and is measured in days. – Ross Millikan May 13 '16 at 14:04
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For the initial mass you only need to go back to when the timing began, so setting t=0 should give you the answer. $$x_0=4e^0=4$$

Ivan Lerner
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