Question

Compound value solving for n) How many years will the following take? a. $490 to grow to $1.270.93 if invested at 10 percent compounded annually b. $36 to grow to $71.73 if invested at 9 percent compounded annually c. $110 to grow to $476.80 if invested at 13 percent compounded annually d. $58 to grow to $77.73 if invested at 5 percent compounded annually a. How many years will t take for $490 to grow to $1,270.93 f invested at 10 percent compounded annually? 10 years (Round to the nearest whole number.) b. How many years willit take for $36 to grow to $71.73 if invested at 9 percent compounded annually? years (Round to the nearest whole number.) Enter your answer in the answer box and then click Check Answer. remaining Clear All

Answer 1

Formula of compound interest implies A = P(1 + r/n)^t middot n where A = Amount: P = principal: r = interest rate t = time: n = no. of time per year, interest is compounded P = $490: A = $1270.93: r = 10% = 0.1 n = 1: t = ? therefore 1270.93 = 190(1 + 0.1)^t (1.1)^t = 1270.93/490 t log (1.1) = log 91270.93/490) = 0.4139255512 t = 0.4132255512/log (1.1) = 9.99 TildeEqual 10(approx): 10year b) p = $36: A = $71.73: r = 9% = 0.09: n = 1 therefore 71.73 = 36(1 + 0.09)^t (1.09)^t = 71.73/36 = 1.9925 log(1.09)^t = log (1.9925) t = log (1.9925)/log (1.09) = 7.99 TildeEqual 8 (approx) \r\n Answer: 8 year c) p = $110: A = $476.80: r = 13% = 0.13: n = 1 therefore 476.80 = 110 (1 + 0.13)^t (1.13)^t = 476.80/110 = 4.3344545455 log (1.13)^t = log (4.334545455)