Date: 03/11/99 at 18:06:19 From: Tom Judd, D.V.M. Subject: Fractional Exponents I am a veterinarian at a large goat farm and I am trying to figure out the water consumption of the average goat. One of the formulae that I have is 145 grams of water per kilogram of bodyweight to the power of .75 (145g per BW Kg^.75). I cannot remember how fractional exponents work! Our average goat weighs 77 Kg. I think 77 kg ^.75 would be (.75 X 77 kg) X (.75 X 77 Kg) = (57.75 X 57.75) = 3335.06 Have I figured this out correctly? Thank you!
Date: 03/15/99 at 12:36:31 From: Doctor Mike Subject: Re: Fractional Exponents A fractional power relates to the expression of the power in numerator and denominator form. You have 3/4 for the 0.75 decimal. Similarly, 0.5 is 1/2 and 0.3333333 is 1/3 and so on. If you represent the fraction as n/d then x to the n/d power means the d-th root of x to the power n. Examples: If d is 2 then we deal with the 2nd root, which is commonly referred to as the square root. If d is 3 then we deal with the 3rd root, which is commonly referred to as the cube root. If d is 4 then we deal with the 4th root, which you get by taking the square root twice. So, for your example of 77 to the 3/4 power, you would first take 77 to the third power, take the square root of that, and then, take the square root again. What we get is SQRT( SQRT( 77*77*77 ) ) = SQRT( SQRT( 456533 ) ) = SQRT( 675.67226 ) = 25.993697 which is about 26. I think, though, that this is a good place to use an 8-dollar pocket calculator. Just enter 77, then the y^x key (or something similar), then 0.75, and finally the = key. Your final answer is 145*25.993697 = 3769.086 grams = 3.769086 kg, or a bit under 4 kilograms of water. Of course, I cannot speak for the accuracy of the formula. I hope this helps. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/
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