# 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.

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.

## 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.

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.
Math10 said:
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.
Math10 said:
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!

## 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.

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).