you invest $1,500 into an account that grows continuously at a rate of 5% per year.\na) write a function…

you invest $1,500 into an account that grows continuously at a rate of 5% per year.\na) write a function that represents the account balance after t years.\nb) how much money will you have after 20 years?\nc) how many years will it take for your investment to reach $10,000? express your answer in logarithmic form and approximate it using a calculator. show work.

you invest $1,500 into an account that grows continuously at a rate of 5% per year.\na) write a function that represents the account balance after t years.\nb) how much money will you have after 20 years?\nc) how many years will it take for your investment to reach $10,000? express your answer in logarithmic form and approximate it using a calculator. show work.

Answer

Explanation:

Step1: Write the continuous - growth formula

The formula for continuous compounding is (A(t)=P\cdot e^{rt}), where (P) is the principal amount, (r) is the annual interest rate (in decimal form), and (t) is the number of years. Given (P = 1500), (r=0.05). So the function is (A(t)=1500e^{0.05t}).

Step2: Calculate the amount after (t = 20) years

Substitute (t = 20) into the formula (A(t)=1500e^{0.05t}). (A(20)=1500e^{0.05\times20}) First, calculate (0.05\times20 = 1). Then (A(20)=1500e^{1}). Using (e\approx2.71828), we have (A(20)=1500\times2.71828 = 4077.42).

Step3: Find the time (t) when (A(t)=10000)

Set (A(t)=10000) in the formula (A(t)=1500e^{0.05t}). (10000 = 1500e^{0.05t}). Divide both sides by (1500): (\frac{10000}{1500}=e^{0.05t}), which simplifies to (\frac{20}{3}=e^{0.05t}). Take the natural logarithm of both sides: (\ln(\frac{20}{3})=\ln(e^{0.05t})). Using the property (\ln(e^{x})=x), we get (\ln(\frac{20}{3}) = 0.05t). Solve for (t): (t=\frac{\ln(\frac{20}{3})}{0.05}). Using a calculator, (\ln(\frac{20}{3})\approx1.8971), then (t=\frac{1.8971}{0.05}=37.942).

Answer:

a) (A(t)=1500e^{0.05t}) b) ($4077.42) c) (t = \frac{\ln(\frac{20}{3})}{0.05}\approx37.94) years