Question 332248


Recall that the formula for continuous compounding interest is


{{{A=Pe^(rt)}}}


where A is the return, P is the principal (amount invested), r is the interest rate (in decimal form) and t is the time in years.



Since "$2000 is invested at a rate of 6% per year compounded continuously", we know that {{{P=2000}}} and {{{r=0.06}}} (the decimal equivalent of 6%).



Now let's compute the return when {{{t=1}}} (ie find the balance after one year)



{{{A=Pe^(rt)}}} Start with the continuous compounding formula.



{{{A=2000*e^(0.06*1)}}} Plug in {{{P=2000}}}, {{{r=0.06}}}, and {{{t=1}}}.



{{{A=2000*e^(0.06)}}} Multiply {{{0.06}}} and {{{1}}} to get {{{0.06}}}.



{{{A=2000*1.06183654654536}}} Raise 'e' (which is approximately 2.71828) to the power {{{0.06}}} to get {{{1.06183654654536}}} (this value is approximate).



{{{A=2123.67309309072}}} Multiply {{{2000}}} and {{{1.06183654654536}}} to get {{{2123.67309309072}}}.



{{{A=2123.67}}} Round to the nearest hundredth (ie to the nearest penny).



So after 1 year, you'll have about $2,123.67



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Now let's compute the return when {{{t=2}}} (ie find the balance after two years)



{{{A=Pe^(rt)}}} Start with the continuous compounding formula.



{{{A=2000*e^(0.06*2)}}} Plug in {{{P=2000}}}, {{{r=0.06}}} (the decimal equivalent of 6%), and {{{t=2}}}.



{{{A=2000*e^(0.12)}}} Multiply {{{0.06}}} and {{{2}}} to get {{{0.12}}}.



{{{A=2000*1.12749685157938}}} Raise 'e' (which is approximately 2.71828) to the power {{{0.12}}} to get {{{1.12749685157938}}} (this value is approximate).



{{{A=2254.99370315875}}} Multiply {{{2000}}} and {{{1.12749685157938}}} to get {{{2254.99370315875}}}.



{{{A=2254.99}}} Round to the nearest hundredth (ie to the nearest penny).



So after two years, you'll have $2,254.99