Friday

December 19, 2014

December 19, 2014

Posted by **Anne** on Thursday, February 9, 2012 at 1:57am.

A(t) = 7000(1 + (0.0394/12))^12t.

At the start of which year will an investment of $7,000 made at the beginning of 2007 first exceed $11,000? (Round your answer the nearest whole number.)

I don't quite understand the question, but multiplying it out I got:

If t=1 then the answer 7280

If t=2 then the answer 14561.7

I don't understand the amount of years. How do I calculate that?

- Math -
**Reiny**, Thursday, February 9, 2012 at 8:33amThe formula finds the amount at the

**end of**so many months

notice the exponent is 12t, so if there are 3 years, t - 3 , but the exponent is 36 (36 months in 3 years)

also notice that the interest rate of .0394 (or 3.94%) is an annual rate, so it is also divided by 12

so let's set A(t) = 11000

11000 = 7000 (1.003283333)^12t

11000/7000 = (1.00328333)^12t

1.003283333^12t = 1.571428571

now take log of both sides and use the log rules

log(1.003283333^12t) = log(1.571428571)

12t log (1.003283333) = log(1.571428571)

12t = log 1.571428.)/(log 1.0032833..) = 137.886 months

or 11.5 years

So we will reach the 11000 after 11.5 years, so according to your question it will have exceeded the 11000 mark at the beginning of year 12.

check

if t = 11 , A(11) = 7000(1.00328333)^132 = 10789.79 , (not there yet)

if t = 12 , A(12) =7000(1.003283333)^144 = 11222.67

btw, you calculation for t = 2 is incorrect

A(2) = 7000(1.00328333)^24 = 7572.94

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