# Math

posted by on .

The associated exponential model for the value of a deposit after t years is
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 - ,

The 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