Find the quantity Q(t) of the substance left?

[Math10]

Homework Statement

The half-life of a radioactive substance is 3200 years. Find the quantity Q(t) of the substance left at time t>0 if Q(0)=20 g.

Homework Equations

dQ/dt=kQ

The Attempt at a Solution

dQ/dt=kQ
dQ/Q=k dt
ln abs(Q)=kt+C
Q=Ce^(kt)
C=20
Q=20e^(kt)
20e^(3200k)=10
e^(3200k)=1/2
Now I'm stucked. The answer is Q=20e^(-(t*ln2)/3200) g.

 

[Mark44]

Homework Statement

The half-life of a radioactive substance is 3200 years. Find the quantity Q(t) of the substance left at time t>0 if Q(0)=20 g.

Homework Equations

dQ/dt=kQ

The Attempt at a Solution

dQ/dt=kQ
dQ/Q=k dt
ln abs(Q)=kt+C

Since the quantity Q can't be negative, there's no need for absolute values.

Q=Ce^(kt)
C=20
Q=20e^(kt)
20e^(3200k)=10
e^(3200k)=1/2

Take the natural log of both sides to solve for k.

Now I'm stucked. The answer is Q=20e^(-(t*ln2)/3200) g.

There is no such word in English as "stucked." You can say that you're stuck.

 

[Math10]

Okay, thank you!

 

[Ray Vickson]

Homework Statement

The half-life of a radioactive substance is 3200 years. Find the quantity Q(t) of the substance left at time t>0 if Q(0)=20 g.

Homework Equations

dQ/dt=kQ

The Attempt at a Solution

dQ/dt=kQ
dQ/Q=k dt
ln abs(Q)=kt+C
Q=Ce^(kt)
C=20
Q=20e^(kt)
20e^(3200k)=10
e^(3200k)=1/2
Now I'm stucked. The answer is Q=20e^(-(t*ln2)/3200) g.

It is a good idea to get in the habit of doing the following:
(1) Write $Q = e^{-kt}$ if $Q = Q(t)$ is decreasing.
(2) Write $Q = e^{kt}$ if $Q = Q(t)$ is increasing.
By doing that, we always have $k > 0$. This is convenient because we usually want to know the magnitude of $k$ (for example, it might be tabulated in a handbook), and then knowing whether or not you should use $+k$ or $-k$ in the exponent is up to the user. (Of course, your method is OK too, but it will give a negative value of $k$.) This sounds like a minor point, but standard usage in most fields adheres to (1) or (2).