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My attempt:

Step 1: Find $x$ in terms of $t$.

  • $\frac{dt}{dx} = \frac{1}{-0.15x}$
  • $t = \frac{1}{-0.15}\ln(x) = x^{-1}(t)$
  • $x(t) = e^{-0.15t}+c$

However, here is where I am stuck. Without any extra information about $x(t)$(initial conditions) to find the value of $c$, I cannot proceed further to find the value of $t$ at which the amount of drug has halved and hence, the time interval as well. Does anyone have any idea of how this could be solved?

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  • $\begingroup$ The notation is a little confusing, which may be why the constant of integration is in the wrong place. Integrating gives $t + C = -\frac{20}{3} \ln x$ for some constant $C$, and exponentiating gives $x(t) = D e^{-3 t / 20}$ for some constant $D = x(0)$, and you'll find that the halving time doesn't depend on the initial amount $x(0)$. $\endgroup$
    – Travis Willse
    Commented May 24, 2022 at 10:39
  • $\begingroup$ Just to be more explicit about what Travis said, you have three substeps in "Step 1". The constant of integration should be added in the SECOND substep, not the third. $\endgroup$
    – johnnyb
    Commented May 24, 2022 at 13:00

1 Answer 1

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That's an awkward way of working on the problem! From dx/dt= -0.15x we can immediately get dx/x= -0.15tdt and, itegrating ln(x)= -0.15t+ C for some constant C. Taking the exponential of both sides, x= C'e^(-0.15t) where C'= e^C.

That seems more natural to me than inverting the equation to "dt/dx" but you got that result anyway!

When t= 0, x= C'. What will t be when x= C'e^(-0.15t) is half that? That is, you need to solve C'e^(-0.15t)= (1/2)C'. Of course, you can divide both sides by C' so your problem is to solve e^(-0.15t)= 1/2.

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