SOLUTION: $3000 is deposited at 8% compounded semiannually. How long does it take to double? Round to the nearest tenth of a year. Use the formula A=P(1+r/n)^nt possible answers 1.) 8.8

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Question 1209221: $3000 is deposited at 8% compounded semiannually. How long does it take to double? Round to the nearest tenth of a year.
Use the formula
A=P(1+r/n)^nt
possible answers
1.) 8.8 years
2.) 9.2 years
3.) 9 years
4.) 8.6 years

Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answer: 8.8 years

Work Shown
P = 3000 is the investment amount. We wish to double it to A = 6000 dollars.
r = 0.08 = decimal form of the annual interest rate
n = 2 since we're compounding 2 times a year
The goal is to solve for t.

A = P(1+r/n)^(nt)
6000 = 3000(1+0.08/2)^(2*t)
6000/3000 = (1.04)^(2*t)
2 = (1.04)^(2*t)
Log(2) = Log( (1.04)^(2*t) )
Log(2) = 2t*Log(1.04)
t = (1/2)*Log(2)/Log(1.04)
t = 8.836494 approximately
t = 8.8 years when rounding to the nearest tenth of a year.

Footnotes:

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

Calculations that brought tutor @math_tutor2020 to the value t = 8.83... years (approximately)
were correct, but the conclusion, which he made from it was wrong.

The discretely compounded amount is not a continuous function of time.

The amount of a discretely compounded account is a discontinuous function.

It is changed jumping at the compounding time moments.

So, the value t= 8.83... years MUST be rounded UP to the closest compounding time period,
in order for the bank would be in position to make the last compounding.

Thus the correct answer to this problem is 9 years.


About a million of similar problems are solved at this site in my lesson
    - Problems on discretely compound accounts
with complete explanations.

Learn the subject from there.


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