# Exponential Function Word Problem

• May 13th 2009, 12:43 PM
skeske1234
Exponential Function Word Problem
Use the decay eqtn for polonium-218, at A(t)=A_o(0.5)^(t/3.1), where A is the amount remaining after t min. and A_o is the initial amount.

a) How much will remain after 90 sec from an initial sample of 50 mg?

b) How long will it take for this sample to decay to 10% of its initial amount of 50 mg?

c) Would your answer in part b change if the size of the initial sample where changed? Explain why or why not.

My answer for part a was 35.75 mg, which is correct.

But I do not understand how to do part b and c. For part b, what I did was sub in 35.75 for A(t) and subbed in 5 (10% of 50, which is 5) for the initial amount A_o. Then, I proceeded on to solve for t.

I ended up with -9.04, but the answer is supposed to be approx 10.3 min. Could someone explain to me what I did wrong and how I can fix this?

Also, for part c, the answer in the back says "no" but I think that (even though I don't have part b done), from the answer that they give, the answer would in fact be "yes" because from part a we used 50 mg as the initial sample and that is how we calculated part b. So they must be the same, otherwise the answer would change in part b. Am I wrong on this too? Please explain why.

Thank you very much to whoever helps!
• May 13th 2009, 01:16 PM
artvandalay11
exponential decay solution
$\displaystyle A_{0}$ is the initial amount in the equation, so you should have plugged in 50 for $\displaystyle A_{0}$

Also, you are right that we have to find out when it decays to 5, but the problem says A is the amount remaining after time t. They in fact mean A(t) is the amount remaining after time t, as the amount remaining is a function of how much you started with. So we need to solve $\displaystyle 5=50(.5)^\frac{t}{3.1}$ and you will get approx 10.3.

For part c) the time it takes to decay to 10% of the original amount actually does not depend on the size of the initial amount. You can see this buy saying let X be the original amount. Then we have to see when it decays to .10*X.... so we solve

$\displaystyle (.1)X=X(.5)^\frac{t}{3.1}$ and you will get approx. 10.3
• May 13th 2009, 01:19 PM
e^(i*pi)
Quote:

Originally Posted by skeske1234
Use the decay eqtn for polonium-218, at A(t)=A_o(0.5)^(t/3.1), where A is the amount remaining after t min. and A_o is the initial amount.

a) How much will remain after 90 sec from an initial sample of 50 mg?

b) How long will it take for this sample to decay to 10% of its initial amount of 50 mg?

c) Would your answer in part b change if the size of the initial sample where changed? Explain why or why not.

My answer for part a was 35.75 mg, which is correct.

But I do not understand how to do part b and c. For part b, what I did was sub in 35.75 for A(t) e^(i*pi): Why did you do this? It's unnecessary, you can follow the following step and solve for t
and subbed in 5 (10% of 50, which is 5) for the initial amount A_o. Then, I proceeded on to solve for t.

I ended up with -9.04, but the answer is supposed to be approx 10.3 min. Could someone explain to me what I did wrong and how I can fix this?

Also, for part c, the answer in the back says "no" but I think that (even though I don't have part b done), from the answer that they give, the answer would in fact be "yes" because from part a we used 50 mg as the initial sample and that is how we calculated part b. So they must be the same, otherwise the answer would change in part b. Am I wrong on this too? Please explain why.

Thank you very much to whoever helps!

$\displaystyle A(t) = A_0(0.5)^{\frac{t}{3.1}}$

b) Let $\displaystyle A(t) = 0.1A_0$

$\displaystyle 0.1A_0 = A_0(0.5)^{\frac{t}{3.1}}$

$\displaystyle 0.1 = (0.5)^{\frac{t}{3.1}}$

take logs to base 10: $\displaystyle -1 = \frac{t}{3.1}log(0.5)$

note that $\displaystyle log(0.5) = log(2^{-1}) = -log(2)$

$\displaystyle \frac{1}{log(2)} = \frac{t}{3.1}$

$\displaystyle t = \frac{3.1}{log(2)} = 10.30$

c) No it wouldn't. Since this is a ratio in question then it would not change. If we set $\displaystyle A(t) = 0.1A_0$ which is saying 10% of the original sample.

Divide that through by $\displaystyle A_0$ and you $\displaystyle A_0$ will cancel thus making it independent of the initial amount
The text passage contains a mathematical word problem involving exponential decay, specifically for polonium-218. The problem has three parts, and the user provides their attempt at solving part (a) and asks for help with parts (b) and (c).
multiple_questions_with_answers

1. the passage contains multiple mathematical questions (parts a, b, and c).
2. answers to these questions are provided in the text, including the user's attempt and the correct answers.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Exponential Function Word Problem

• May 13th 2009, 12:43 PM
skeske1234
Exponential Function Word Problem
Use the decay eqtn for polonium-218, at A(t)=A_o(0.5)^(t/3.1), where A is the amount remaining after t min. and A_o is the initial amount.

a) How much will remain after 90 sec from an initial sample of 50 mg?

b) How long will it take for this sample to decay to 10% of its initial amount of 50 mg?

c) Would your answer in part b change if the size of the initial sample where changed? Explain why or why not.

My answer for part a was 35.75 mg, which is correct.

But I do not understand how to do part b and c. For part b, what I did was sub in 35.75 for A(t) and subbed in 5 (10% of 50, which is 5) for the initial amount A_o. Then, I proceeded on to solve for t.

I ended up with -9.04, but the answer is supposed to be approx 10.3 min. Could someone explain to me what I did wrong and how I can fix this?

Also, for part c, the answer in the back says "no" but I think that (even though I don't have part b done), from the answer that they give, the answer would in fact be "yes" because from part a we used 50 mg as the initial sample and that is how we calculated part b. So they must be the same, otherwise the answer would change in part b. Am I wrong on this too? Please explain why.

Thank you very much to whoever helps!
• May 13th 2009, 01:16 PM
artvandalay11
exponential decay solution
$\displaystyle A_{0}$ is the initial amount in the equation, so you should have plugged in 50 for $\displaystyle A_{0}$

Also, you are right that we have to find out when it decays to 5, but the problem says A is the amount remaining after time t. They in fact mean A(t) is the amount remaining after time t, as the amount remaining is a function of how much you started with. So we need to solve $\displaystyle 5=50(.5)^\frac{t}{3.1}$ and you will get approx 10.3.

For part c) the time it takes to decay to 10% of the original amount actually does not depend on the size of the initial amount. You can see this buy saying let X be the original amount. Then we have to see when it decays to .10*X.... so we solve

$\displaystyle (.1)X=X(.5)^\frac{t}{3.1}$ and you will get approx. 10.3
• May 13th 2009, 01:19 PM
e^(i*pi)
Quote:

Originally Posted by skeske1234
Use the decay eqtn for polonium-218, at A(t)=A_o(0.5)^(t/3.1), where A is the amount remaining after t min. and A_o is the initial amount.

a) How much will remain after 90 sec from an initial sample of 50 mg?

b) How long will it take for this sample to decay to 10% of its initial amount of 50 mg?

c) Would your answer in part b change if the size of the initial sample where changed? Explain why or why not.

My answer for part a was 35.75 mg, which is correct.

But I do not understand how to do part b and c. For part b, what I did was sub in 35.75 for A(t) e^(i*pi): Why did you do this? It's unnecessary, you can follow the following step and solve for t
and subbed in 5 (10% of 50, which is 5) for the initial amount A_o. Then, I proceeded on to solve for t.

I ended up with -9.04, but the answer is supposed to be approx 10.3 min. Could someone explain to me what I did wrong and how I can fix this?

Also, for part c, the answer in the back says "no" but I think that (even though I don't have part b done), from the answer that they give, the answer would in fact be "yes" because from part a we used 50 mg as the initial sample and that is how we calculated part b. So they must be the same, otherwise the answer would change in part b. Am I wrong on this too? Please explain why.

Thank you very much to whoever helps!

$\displaystyle A(t) = A_0(0.5)^{\frac{t}{3.1}}$

b) Let $\displaystyle A(t) = 0.1A_0$

$\displaystyle 0.1A_0 = A_0(0.5)^{\frac{t}{3.1}}$

$\displaystyle 0.1 = (0.5)^{\frac{t}{3.1}}$

take logs to base 10: $\displaystyle -1 = \frac{t}{3.1}log(0.5)$

note that $\displaystyle log(0.5) = log(2^{-1}) = -log(2)$

$\displaystyle \frac{1}{log(2)} = \frac{t}{3.1}$

$\displaystyle t = \frac{3.1}{log(2)} = 10.30$

c) No it wouldn't. Since this is a ratio in question then it would not change. If we set $\displaystyle A(t) = 0.1A_0$ which is saying 10% of the original sample.

Divide that through by $\displaystyle A_0$ and you $\displaystyle A_0$ will cancel thus making it independent of the initial amount
user
The text passage contains a mathematical word problem involving exponential decay, specifically for polonium-218. The problem has three parts, and the user provides their attempt at solving part (a) and asks for help with parts (b) and (c).

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

1. The passage contains multiple mathematical questions (parts a, b, and c).
2. Answers to these questions are provided in the text, including the user's attempt and the correct answers.