# uniqueness of log

• Feb 21st 2009, 10:45 AM
manjohn12
uniqueness of log
Fix $b>1$ and $y>0$. Prove that there is a unique real $x$ such that $b^{x} = y$.

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

So for any positive integer $n$, $b^{n}-1 \geq n(b-1)$. We can show this by factoring the LHS as $(b-1)(b^{n-1}+b^{n-2}+b^{n-3} + \ldots + b^{2} + b+1) \geq (b-1)n$. So $b-1 \leq \frac{b^{n}-1}{n}$. Or $b-1 \geq n(b^{1/n}-1)$. From here what do we do?
• Feb 21st 2009, 11:22 AM
manjohn12
In other words the strategy is that we need to consider a set....show that $x$ is the supremum of that set...and then show uniqueness?
• Feb 21st 2009, 02:13 PM
manjohn12
See what is the point of showing that is is strictly increasing? I see. Because you can show uniqueness by exploiting monotonicity.
• Feb 21st 2009, 02:49 PM
manjohn12
Let us consider the slightly different problem of showing that there is a unique $b$. So we fix $x, y>0$. Then uniqueness follows because $0 < y_1 < y_2 \implies y_{1}^{n} < y_{2}^{n}$. And to show existence we consider the set of $t$'s such that $t^n < x$.

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