base e: making negative exponents positive
in [Math]
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From: davidcsnow on 8 Feb 2005 16:40 is the following possibe? and if so, why? I havent looked at this stuff in a while, so its a bit confusing...and help would be great! Start with: e (to the power of) -5/t = 0.20 Then I have to change '-5/t' to '+5/t' giving: e (to the power of) +5/t = 5.00 I just dont understand, and cant find the rules that explain why 0.20 becomes 5.00 when you make the negative exponent positive. Can anyone explain this to me? Thanks so much in advance, David.
From: Larry Lard on 8 Feb 2005 16:53 davidcsnow(a)yahoo.com wrote: > is the following possibe? and if so, why? I havent looked at this stuff > in a while, so its a bit confusing...and help would be great! > > Start with: > > e (to the power of) -5/t = 0.20 Use ^ for (to the power of) in ASCII text. > > Then I have to change '-5/t' to '+5/t' giving: > > e (to the power of) +5/t = 5.00 > > I just dont understand, and cant find the rules that explain why 0.20 > becomes 5.00 when you make the negative exponent positive. > > Can anyone explain this to me? What does 2 ^ -3 mean? How do you evaluate it? -- Larry Lard Replies to group please
From: Michael Stemper on 8 Feb 2005 17:07 In article <1107898827.565319.211470(a)l41g2000cwc.googlegroups.com>, davidcsnow writes: >is the following possibe? and if so, why? I havent looked at this stuff >in a while, so its a bit confusing...and help would be great! > >Start with: > >e (to the power of) -5/t = 0.20 > >Then I have to change '-5/t' to '+5/t' giving: > >e (to the power of) +5/t = 5.00 > >I just dont understand, and cant find the rules that explain why 0.20 >becomes 5.00 when you make the negative exponent positive. It's the laws of exponents. Let's start with integer exponents. (Keep in mind that "x squared" is usually written as "x^2".) x^3 = x*x*x, right? (x^3)/x = x^2 = x*x, right? (x^2)/x = x^1 = x, right? This is pretty simple so far (I hope). To keep things consistent, mathematicians then defined (x^1)/x = x^0 = 1. Do you see that this makes some sense? Each time that we divide by x, we reduce the exponent by one. The next law of exponents to think about is that x^(a+b) = (x^a)*(x^b). For instance, x^3 = x^(1+2) = (x^1)*(x^2) = x*x*x. Do you buy that? If so, then to keep things consistent, define: x^(-y) = 1/(x^y) Why is this consistent? Because if we take (x^y)*(x^(-y)), we get x^(y+(-y)), or x^0, which is 1. So, changing the sign of the exponent *means* taking the reciprocal. (One other note: everything that I've said above only works for positive values of x. The exponents can be positive, negative, zero, integer, rational, or real. The base has to be positive, or things get much more complicated. In your example, the base was "e", which is a positive number, so we're OK.) -- Michael F. Stemper #include <Standard_Disclaimer> Visualize whirled peas!
From: davidcsnow on 8 Feb 2005 17:34
Got it, so because the exponent is -ve, you take the reciprocal. You then have to evaluate the RHS of the equation, which is 1/0.2, equalling 5. Superb, thx a lot... |