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Does the logarithmic function of base $2$ have a supremum? Just wondering because the $\log_2 (10^{135})$ is $448.460293$ and $\log_2 (10^{164})$ is $544.796208$

If it does not exist, can you prove it?

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Good that you computed. In the long run, $\log_2$ grows glacially slowly, but it gets arbitrarily large. –  André Nicolas Dec 9 '12 at 18:01

3 Answers 3

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If we wish to show that there is no real supremum, then we must show that for all $M>0$, there exists $n>0$ such that $\log_2(x)> M$ for all $x\geq n$. In particular, noting that $\log_2(x)$ is an increasing function on $\Bbb R^+$, we need only find some $n>0$ such that $\log_2(n)> M$, from which the rest will follow. Indeed, putting $n>2^M$, we have $$\log_2(n)>\log_2(2^M)=M.$$ Thus, the sequence $\{\log_2(n)\}_{n=1}^\infty$ has no upper bound in the reals, and so no least upper bound (supremum) in the reals.

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No, it hasn't because $\log_2(2^n)=n$, in particular for every natural $n$ which you can choose arbitrarily large.

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I think what you are trying to ask is whether or not $\log_{2}(x)$ has a limit. By definition, it is the inverse of the exponential function $2^{x}$.
If there was some natural $M$ such that $\log_{2}(x)<M$ for all $x$, then this would mean also that $x<2^{M}$ for all $x$ - in other words, there is a largest natural number. What does this tell you?

What you've noticed is that it grows very slowly, which is surely true, This is a consequence of how fast $2^{x}$ grows. Can you see why?

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