SOLUTION: 1. You have deposited $7,000 in an account that pays 6.25% interest compounded continuously. How long will it take your money to double? 2. You have deposited $12,000 in an

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: 1. You have deposited $7,000 in an account that pays 6.25% interest compounded continuously. How long will it take your money to double? 2. You have deposited $12,000 in an      Log On

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Question 929310: 1. You have deposited $7,000 in an account that pays 6.25% interest compounded continuously. How long will it take your money to double?

2. You have deposited $12,000 in an account compounded monthly. After 7 years the balance has grown to $15,532. Find the rate.

PLEASE HELP ME ANSWER THESE 2 QUESTIONS!!!

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1. For continuous compounding, the balance amount A is related to
the principal (initial investment P ,
the rate r (written as a decimal), and
the time in years, t , by the formula
A=P%2Ae%5E%28%28r%2At%29%29 .
The number e is an irrational number like pi that come up pretty often . It is not invented to make students learn one more number; it just comes up naturally.
We use it as a base for powers, and of course for logarithms that we call natural logarithms
6.25%25=6.25%2F100=0.0625
So, for P=7000 , r=0.0625 , and A=2%2AP=2%2A7000 , we can write
2%2AP=P%2Ae%5E0.0625t ---> 2=e%5E0.0625t
Note that we did not need to know the amount invested.
2=e%5E0.0625t ---> ln%282%29=0.0625t (taking natural logarithms on both sides).
Using the approximate value ln%282%29=0.693147 ,
0.0625t=0.693147-->t=0.693147%2F.0625-->t=11.09 (rounded).
So, it will take 11 years for the money to double.
2. $12,000 compounded monthly. After 7 years the balance has grown to $15,532. Find the rate.
For monthly compounding, the balance grow by %281%2F12%29 of the yearly rate every month.
So over 7years=7%2A12months=84months, the balance grew by a factor of 1%2B%28rate%2F12%29 applied 84 times to
15320=12000%2A%281%2Brate%2F12%29%5E84 .
Solving for rate:
15320=12000%2A%281%2Brate%2F12%29%5E84
15320%2F12000=%281%2Brate%2F12%29%5E84
1.2766666667=%281%2Brate%2F12%29%5E84
log%281.2766666667%29=84%2Alog%281%2Brate%2F12%29
%281%2F84%29%2Alog%281.2766666667%29=log%281%2Brate%2F12%29
Using logarithm base 10 for an approximate calculation:
%281%2F84%29%2A0.10677519=log%281%2Brate%2F12%29
0.0012628=log%281%2Brate%2F12%29
Reversing the logarithm by calculating power of 10 on both sides of the equal sign:
10%5E0.0012628=1%2Brate%2F12
1.002912=1%2Brate%2F12
1.002912-1=rate%2F12
0.002912=rate%2F12
rate=0.03496=%223.494%25%22