can you love this question for me please An engineer borrows

can you love this question for me please

An engineer borrows $10 000 lo buy a personal computer. Me must repay $218.94 a month for 5 years What is the nominal annual interest rate, based upon continuous compounding?

Solution

The formula for continuous compounding is A = Pe^{rt ,} where P is the initial amount, t is the number of years, A is the amount after time t years, and r is the annual rate of interest in decimals. Here, P = $ 10000, t = 5 , and A = $ 218.94*12*5 = $ 13136.40 ( $ 218.94 per month means $ 218.94*12 = $ 2627.28 p.a.. This translates into $  2627.28* 5 = $ 13136.40 in 5 years) . Let the rate of interest be r % per annum (year). Then , we have 13136.40 = 10000*e^5r. Therefore, e^5r = 13136.40 / 10000 = 1.31364. On taling natural logarithms of both the sides, we get ln(  e^5r ) = ln (1.31364) or, 5r ( ln e ) = ln (1.31364) or, 5r = 0.272801909. Therefore, r = 0.272801909/ 5* 100 % or, r = 5.456038198 % = 5.46 % ( on rounding off to 2 decimal places).

We can verify this result by working backwards. Converting the rate of interest in decimals ( it is in percentage now), we have r = 5.456038198 /100 = 0. 05456038198 . Now, rt = 5* 0.05456038198 =  0.272801909 . Finally,

e^0.272801909 = 1.31364. Therefore, Pe^rt =  10000* e^0.272801909 = 10000*1.31364.= 13136.40.