Question

If a quantity of substance decays exponentially, the following equation relates the amount of time t it takes for the substance to lose half its value to the decay rate k:
12Av0=Av0e^kt.
a. Solve the equation for t in terms of k.
b. What is the half-life of a substance whose decay rate is -7% per year?
Round your answer to the nearest tenth of a year.
c. Solve the equation for k in terms of t.
d. If the half-life of a substance is 7 years, find its decay rate. Round your answer to nearest tenth of a percent.

Answer 1

@Isaiah.Feynman

Answer 2

\[\frac{ 1 }{ 2 }A_{0} = A_{0}e^{kt}\]

Answer 3

Since we are dealing with half lives, that models simply says at some initial time (A_0), a substance has a certain quantity, after it stays past its half life, the new quantity becomes half of the original quantity.

Answer 4

In the book, it said it found k before they found t.

Answer 5

Okay. I was just explaining the model in your question in words so its easier.

Answer 6

Okay, so what's next?

Answer 7

We answer the questions. To solve for t in terms of k.

Answer 8

We cancel out A to get \[\frac{ 1 }{ 2 } = e^{kt}\]

Answer 9

It says there is no solution when I plug in -0.07 for k

Answer 10

Okay. We haven't even answered the first question though.

Answer 11

I'm getting no solution for that as well.

Answer 12

Hang on.

Answer 13

So, we use the natural logarithm to bring the t. Like so \[\ln \frac{ 1 }{ 2 } = kt \]

Answer 14

So, \[t = \frac{\ln \frac{ 1 }{ 2 } }{ k }\]

Answer 15

Then we plug -0.07 in k?

Answer 16

We just answered the first question.

Answer 17

I found out something that's wrong with the question. If its exponential decay, then the formula should look like this \[\frac{ 1 }{ 2 }A _{0} = A_{0}e^{-kt}\]

Answer 18

Which means...\[t = \frac{\ln \frac{ 1 }{ 2 } }{ -k }\]

Answer 19

For the second question, the half life if the decay rate is -7% is...\[t = \frac{ \ln\frac{ 1 }{ 2 } }{ -(-0.07) }\] Since k = -0.07