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Thread: uniqueness of log

  1. #1
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    uniqueness of log

    Fix $\displaystyle b>1 $ and $\displaystyle y>0 $. Prove that there is a unique real $\displaystyle x $ such that $\displaystyle b^{x} = y $.

    So to show uniqueness we need to consider the following: if $\displaystyle b^{x} = y $ and $\displaystyle b^{l} = y $ then $\displaystyle y = l $. But first we want to show that $\displaystyle b^{x} $ is monotonically increasing.

    So for any positive integer $\displaystyle n $, $\displaystyle b^{n}-1 \geq n(b-1) $. We can show this by factoring the LHS as $\displaystyle (b-1)(b^{n-1}+b^{n-2}+b^{n-3} + \ldots + b^{2} + b+1) \geq (b-1)n $. So $\displaystyle b-1 \leq \frac{b^{n}-1}{n} $. Or $\displaystyle b-1 \geq n(b^{1/n}-1) $. From here what do we do?
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  2. #2
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    In other words the strategy is that we need to consider a set....show that $\displaystyle x $ is the supremum of that set...and then show uniqueness?
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  3. #3
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    See what is the point of showing that is is strictly increasing? I see. Because you can show uniqueness by exploiting monotonicity.
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  4. #4
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    Let us consider the slightly different problem of showing that there is a unique $\displaystyle b $. So we fix $\displaystyle x, y>0 $. Then uniqueness follows because $\displaystyle 0 < y_1 < y_2 \implies y_{1}^{n} < y_{2}^{n} $. And to show existence we consider the set of $\displaystyle t $'s such that $\displaystyle t^n < x $.

    So we use this same set in our problem. But we are trying to use this strategy: $\displaystyle \text{monotonic increasing} \implies \text{uniqueness} $.
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