word problems

J

jokerman

New member
Joined
Jun 11, 2007
Messages
13
1. Suppose that you invest $2000 into an account that pays 4.3% interest compounded continuously, find to the tenth of a year when the account will be worth $10,000.

2. Suppose you have 500 grams of a radioactive element who half-life is 27 years.
a) Find the function for the quantity, Q, left in t years.

b) Find to the nearest tenth of a year when there will be 100 grams left.
 
SOOOOOO.....how far did you get?
Were you given formulas and stuff?
 
Okay, so the amount of interest P at any time t (in years) since the investment is: \(\displaystyle P = 2000(1.043)^{t}\). The account will be worth $10, 000 dollars when P = 10, 000, so you need to solve: \(\displaystyle 10 000 = 2, 000(1.043)^{t}\). If you divide both sides by 2000, it simplifies it significantly: \(\displaystyle 5 = (1.043)^{t}\).

Okay, the next one, it basically tells you that 27 years after the element starts to decay, it will have reached 250 (half of its amount) grams. Let's assume it decreases exponentially at r%, then, Q (in grams, t years since it starts to decay), is:

\(\displaystyle Q = 500(\frac{(100 - r)}{100}\)^{t}\)

Now, we know that when t = 27, Q = 250:

\(\displaystyle 250 = 500(\frac{(100 - r)}{100}\)^{27})\), so: \(\displaystyle 1 = 2(\frac{(100 - r)}{100}\)^{27}\)

Then I guess you'd solve for r. Then, once you've found Q in terms of t only, plug Q = 100. By the way: you can shortcut the process and solve for \(\displaystyle \frac{(100 - r)}{100}\\).
 
can you explain a bit more to me because honestly im more confused now then when i started, if you can just explain a bit more and tell me the solutions to the questions i asked it would help me the most, i just need to see how to work the problems i gave to the solution, once i see it i'll know how to do it on similar future problems but right now im getting confused trying to understand what you did on both of them. :(
 
jokerman said:
im more confused now then when i started...
Click to expand...
At what point does the step-by-step explanation stop making sense to you? You are familiar with exponentials and logs, right?

Please be specific. Thank you! :D

Eliz.
 
jokerman said:
1. Suppose that you invest $2000 into an account that pays 4.3% interest compounded continuously, find to the tenth of a year when the account will be worth $10,000.
Click to expand...

The formula for continuous compounding is

A = P e<SUP>rt</SUP>

where A is the ending amount, P is the amount invested, r is the interest rate expressed as a decimal, and t is the number of years.

So, you'll have

10000 = 2000*e<SUP>0.043*t</SUP>

Divide both sides by 2000:

5 = e<SUP>0.043t</SUP>

Take the natural log of both sides:

ln 5 = ln e<SUP>0.043t</SUP>

ln 5 = 0.043t

Divide both sides by 0.043.....

(ln 5) / 0.043 = t

37.4 = t

(I rounded to the nearest tenth as the directions specified)
 
Did you learn anything? Will you be able to solve the next problem on your own?
 
You must log in or register to reply here.
Top