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Math Help - 1=x^0 can x have a unique solution?

  1. #1
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    1=x^0 can x have a unique solution?

    If I have 1=(1+r)^t and I am given t=0, then any r will satisfy the equation, so r is undefined.

    If I re-arrange the formula and have 1+r= 1/(1^t) and am then given t=0 it would appear that there is a single solution of r=0.

    How can I reconcile these two thoughts?
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  2. #2
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    Re: 1=x^0 can x have a unique solution?

    Quote Originally Posted by vicolivo1 View Post
    If I have 1=(1+r)^t and I am given t=0, then any r will satisfy the equation, so r is undefined.

    If I re-arrange the formula and have 1+r= 1/(1^t) and am then given t=0 it would appear that there is a single solution of r=0.

    How can I reconcile these two thoughts?
    You have 1 + r = 1^{1/t}, which is not defined if t = 0 (so the substitution t = 0 into it is not valid).
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    Re: 1=x^0 can x have a unique solution?

    Quote Originally Posted by vicolivo1 View Post
    If I have 1=(1+r)^t and I am given t=0, then any r will satisfy the equation, so r is undefined.

    If I re-arrange the formula and have 1+r= 1/(1^t)
    No, you don't. You get 1+ r= 1^{1/t} which is NOT the same as 1^{-t}= 1/1^t

    and am then given t=0 it would appear that there is a single solution of r=0.

    How can I reconcile these two thoughts?
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