- #1
tronter
- 185
- 1
Fix [itex] b >1, \ y >0 [/itex], and prove that there is a unique real [itex] x [/itex] such that [itex] b^{x} = y [/itex].
Here is the outline:
(a) For any positive integer [itex] n [/itex], [itex] b^{n}-1 \geq n(b-1) [/itex]. Why do we do this?
(b) So [itex] b-1 > n(b^{1/n}-1) [/itex].
(c) If [itex] t>1 [/itex] and [itex] n > (b-1)/(t-1) [/itex] then [itex] b^{1/n} < t [/itex].
etc..
Is this the correct process?
Here is the outline:
(a) For any positive integer [itex] n [/itex], [itex] b^{n}-1 \geq n(b-1) [/itex]. Why do we do this?
(b) So [itex] b-1 > n(b^{1/n}-1) [/itex].
(c) If [itex] t>1 [/itex] and [itex] n > (b-1)/(t-1) [/itex] then [itex] b^{1/n} < t [/itex].
etc..
Is this the correct process?