Question

Suppose that $16,416 is invested at an interest rate of 5.5% per year, compounded continuously. a) Find the exponential function that describes the amount in the account after time t, in years. b) What is the balance after 1 year? 2 years? 5 years? 10 years? c) What is the doubling time? a) The exponential growth function is P(t) = (Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any

Answer 1

Snittal amount (Po) =$16.416 interest rate (P) = 5.5% = 5.5 To let time -t = 0.055 aud the final amount =PLt Then the contineas compound interest formula P(t) = Po e 0.055+ 7 Plt) = 16 416 e The exponential growth timetion is P(t) = 116116e pl is given by, [16116 0.055€ A १ When t=1 0.055X1 Plt) = 16416 e - 17 344-17073 = 17344:17 (too desimal places) when t=2, 0.055X2 Plt) 16416.e 18 324.8208 = 18394.82 ( todo decimal places 3) When t=5 0.055*5 P(t = 16416e 21612.16 75% = 34612.17 ( tooo decimal places)

47 When t=10 0.053 X10 Plt) = 16416. e = 28452.08154 = 28453.08 (teco decimal places) the balace aften 1 year $17314.17 2 years $18394.82 5 year $121612:17 10 $128453.08 u u u и ue u ey e = 2 When Plt) = 16416*2 then t=3 then from contineous compound formula We have 0-055 Xt Plt)? z Poe 0.055t 16416 X 23 16416.e o-Osst taking natural log on both side 0.055t ene = ln 2 [. " ena M Minx => 0.055. E z lna me=1 i t 112 = 12.602 0.055 = 1206 cone decimal places Hem ee to doubting time = 12:6] year.