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?
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
To calculate the number of years it will take for the investment to first exceed $11,000, you need to solve the equation A(t) = $11,000 for t.
Given the exponential model A(t) = 7000(1 + (0.0394/12))^12t, you can substitute A(t) with $11,000 in the equation and solve for t.
11000 = 7000(1 + (0.0394/12))^12t
To solve for t, you need to isolate the variable. Start by dividing both sides of the equation by 7000:
11000/7000 = (1 + (0.0394/12))^12t
1.5714 = (1 + (0.0394/12))^12t
Next, take the natural logarithm (ln) of both sides of the equation to eliminate the exponential function:
ln(1.5714) = ln((1 + (0.0394/12))^12t)
Using the property ln(a^b) = b * ln(a), the equation becomes:
ln(1.5714) = 12t * ln(1 + (0.0394/12))
Divide both sides of the equation by 12 * ln(1 + (0.0394/12)):
ln(1.5714) / (12 * ln(1 + (0.0394/12))) = t
Now, you can calculate t by plugging the values into a calculator. The result for t will give you the number of years it will take for the investment to first exceed $11,000.
Finally, round the value of t to the nearest whole number, as instructed in the question, to determine at the start of which year the investment will exceed $11,000.