Question

At the Beginning of 2001, Clovis invested $10,000 in an account. In 2013, his investment was worth $14,764.
Isobel made an investment at the same time as Clovis. Her investment doubles every 7 years. In 2014, Clovis had 8 times as much in his account as Isobel had in hers.
Assume the values of both investments are exponential functions of time
take t=0 in 2001.

What was the value of Isobel's investment at the beginning of 2001?
pre-calc. problem help!!!

Answer 1

I'll come to your question in just a sec. Basically, you first need to find the correct function for Clovis, then find the amount Clovis has in 2014. Then you need to find the initial value for Isobel.

Answer 2

clovis has about $15,251.206 in 2014

Answer 3

Isobel has about $1906.401 in 2014

Answer 4

Okay, now I'm not sure if those values are correct, but you can proceed from there by finding the function for Isobel, which I'll call I(t)

So as you said, in 2014, Isobel has 1906.401. We can thus say that \(I(13)=1906.401\). We can plug that in her function..
\(I(t)=a(2)^{\frac{t}{7}}\)
You know that \(I(13)=1906.401\), so plug that point in and find a using your logarithms.

Answer 5

Yes, those values that you found are what I got

Answer 6

And the a value you get is the answer because they start at the same time.

Answer 7

so is a 526.209?

Answer 8

I did 1906.401= a(2)^13/7

Answer 9

Yeah that is what I got as well. :)

Answer 10

thx, just didn't know what to do

Answer 11

All good :) just comes with practice.