Just testing myself with some tricky questions in my further maths textbook. This one states that 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
First I am asked what the mass was at the start of the timing. I've established that I can just alter the exponential to 0, because e^0 = 1 Therefore this answer is 4kg.
Next I am asked to find the mass after five years. Am I on the right lines by saying that because the timing is the same (nothing new is specified), I can insert 5 into the timing variable and then multiply this by 365? So 4e^5 x 265
The last question is stumping me a bit. I am being asked to find the time taken for the mass remaining to be half of the initial mass. So I assume I am to take the mass after 5 days (at the end of question 2) and compare it to the initial mass (question 1). Though it says to find the time taken for it to be half the initial mass (which is 2kg, because the ans to question 1 is 4kg). What is the best mathematical way to arrive at this answer?
Thanks for you help. Sorry I haven't done it all myself.
Assuming the actual function is
$$x(t) = 4e^{-0.2t}$$
We are asked to find the half-life time, i.e., the point in time $t$ where $$x(t) = \frac{x(0)}{2}$$
Plugin in above equation gives us
$$4e^{-0.2t}= \frac{4e^{0}}{2}$$ $$4e^{-0.2t}= \frac{4}{2}$$ $$4e^{-0.2t}= 2$$ $$e^{-0.2t}= \frac{1}{2}$$
Now we take the $\ln$ logarithm of both sides, since on the left side we have an exponential with base $e$.
$$e^{-0.2t}= \frac{1}{2}$$ $$-0.2t= \ln\left(\frac{1}{2}\right)$$ $$t= \frac{\ln\left(\frac{1}{2}\right)}{-0.2}$$ $$t = 3.4657359...$$
So if we plug this value in, we get
$$x(3.4657359..) = 2$$
Which is the answer we wanted. After $t = 3.4657359$, the mass is only $2$ kg.