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?

Re: 1=x^0 can x have a unique solution?

Quote:

Originally Posted by

**vicolivo1** 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 $\displaystyle 1 + r = 1^{1/t}$, which is not defined if t = 0 (so the substitution t = 0 into it is not valid).

Re: 1=x^0 can x have a unique solution?

Quote:

Originally Posted by

**vicolivo1** 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 $\displaystyle 1+ r= 1^{1/t}$ which is NOT the same as $\displaystyle 1^{-t}= 1/1^t$

Quote:

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?