What is the most efficient way to solve a problem
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- Is there an efficient way to solve this problem?
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- What is the most efficient method of solving this mathematical problem? 25 X ?^50 = 1300?
- Q: I am trying to work out the the interest rate required to turn £25 in 1300 in 50 years. The interest is compounded so please dont simply post the average rate. I know there is a simple method for calculating this but cannot remember what it is!Many Thanks
- A: what's the compounding interval? the general compound interest formula is:1300 = 25(1 + r/n) ^(50n) (where n is the number of compounding intervals per year)assuming that n = 1, then1300 = 25(1 + r)^(50)52 = (1 + r)^50tak the 50th root of both sides:52^(1/50) = 1 + rr = 52^(1/50) - 1r will be the annual interest rate as a decimal-- .0822this corresponds to 8.22% APRif it compounds more than once a year, just replace n with the number of compounding intervals ex: daily compounding ==> n = 3651300 = 25(1 + r/365)^(50 * 365)52 = (1 + r/365) ^(50 * 365) since 50 * 365 = 18250, you'll have to take the 18250th root of both sides52^(1/18250) = 1 + r/365now subtract 152^(1/18250) - 1 = r / 365and multiply by 365365(52^(1/18250) - 1) = .079, or 7.90%the daily compounding allows for a slightly lower APR to achieve the goaland what the heck...for continuous compounding:1300 = 25e^(50r)52 = e^(50r)okay, now you do need logs!ln 52 = 50rr = (ln 52)/50r = .0790, or 7.90%-- not a whole log of difference between daily and continuous compounding over those 50 years!
- Why can't we solve the high gas price problem with more efficient cars? Surely engineers can design them.
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