SOLUTION: what interest rate compounded daily is required for a $22,000 investment to grow to $36,500 in 5 years?

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Question 391699: what interest rate compounded daily is required for a $22,000 investment to grow to $36,500 in 5 years?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

A=P%281%2Br%2Fn%29%5E%28n%2At%29 Start with the compound interest formula


36000=22000%281%2Br%2F365%29%5E%28365%2A5%29 Plug in A=36000, P=22000, n=365 and t=5.


36000=22000%281%2Br%2F365%29%5E%281825%29 Multiply 365 and 5 to get 1825.


36000%2F22000=%281%2Br%2F365%29%5E%281825%29 Divide both sides by 22000.


1.63636363636364=%281%2Br%2F365%29%5E%281825%29 Evaluate 36000%2F22000 to get 1.63636363636364.


root%281825%2C1.63636363636364%29=1%2Br%2F365 Take the 1825th root of both sides.


1.00026988654164=1%2Br%2F365 Take the 1825th root of 1.63636363636364 to get 1.00026988654164.


1.00026988654164-1=r%2F365 Subtract 1 from both sides.


0.00026988654164195=r%2F365 Combine like terms.


365%2A0.00026988654164195=r Multiply boths sides by 365 to isolate "r".


0.0985085876993119=r Multiply 365 and 0.00026988654164195 to get 0.0985085876993119.


r=0.0985085876993119 Rearrange the equation.


r=0.0985 Round to the nearest ten-thousandth.


So the interest rate is 9.85% (multiply by 100 to convert to a percentage)


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim