Question

A radioactive substance decays at a continuous rate of 2% per year. Suppose that 5 grams are present as of now.

a) how many grams will be present after 10 years?

b) find the half-life.

Answer 1

\[
a(t)=a_0(100\%-2\%)^t
\]

Answer 2

First of all, when they say decrease by \(2\%\), that tells you that the new percent it will be is \(100\%-2\%\).

Answer 3

ok so 98%

Answer 4

Which evaluates to \(0.98\).

Answer 5

When \(t=0\), then it will evaluate to \(1\). So that means \(a_0\) must be the initial amount.

Answer 6

So I guess, we can plug that in: \[
a(t) = 5(0.98)^t
\]

Answer 7

And they want you to find \(a(10)\) for the first part.

Answer 8

For the second part, with the half life we want to set \(a(t) = a_0/2\), that is, half the initial amount. Then we can solve for \(t\).

Answer 9

So that leaves us with: \[
2.5=5(9.8)^t
\]Solving for \(t\).

Answer 10

Am I going to fast, or do you think you got it now?

Answer 11

so for the first part would i solve it by calculating \[5(0.98)^{10}\] ?

Answer 12

Yes

Answer 13

and the second part the same thing?

Answer 14

Second part is solving for \(t\).

Answer 15

can you help explain how i solve for t for the second part?

Answer 16

...

Answer 17

Logarithms.

Answer 18

this is the part i get confused on...

Answer 19

Okay, first: \[
2.5=5(0.98)^t
\]What is the first thing you want to do?

Answer 20

divide from both sides?

Answer 21

Yes, divide by \(5\) first.

Answer 22

.5

Answer 23

Okay, so \[
\frac{1}{2}=(0.98)^t
\]

Answer 24

Now do natural log of both sides: \[
\ln(1/2) = \ln(0.98^t)
\]

Answer 25

Do you know how to simplify this?

Answer 26

ln(1/2) divided by ln(0.98)?

Answer 27

Yes.

Answer 28

34.30961849?

Answer 29

Yes, that approximate.

Answer 30

whats that rounded to the nearest hundredth?