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Since the logarithm is the reversed exponentiation, why does it need a distinct notation for it? Why can't we just ask:


Instead of:

$$\log_2 8=x$$

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For the same reason you write $x=\sqrt{25}$ when you solve $x^2=25$. You want to have an explicit notation for solutions of numeric prblems whenever possible. – Andrea Mori Aug 28 '12 at 21:03
You don't need to, its just awfully convenient. – copper.hat Aug 28 '12 at 21:04
Why the downvotes? I think that is an excellent question. – Michael Greinecker Aug 28 '12 at 21:27
see too this thread – Raymond Manzoni Aug 28 '12 at 22:38
Yes, why bother with multiplication when division will suffice? Or addition even, everyone knows x + y is really x - (0 - y). – Kevin Aug 29 '12 at 1:26
up vote 26 down vote accepted

Suppose you would like to express the fact that, say, $$\lim_{n\to\infty} \left(-\log_e n + \sum_{i=1}^n \frac1i\right) = 0.577\ldots.$$ How do you propose to do this with no $\log$ notation?

Here is another example. Suppose we have a communications channel—say, a telephone cable—over which we can transmit $C$ bits per second.

We want to use this cable to send a sequence of messages, but don't know ahead of time what messages we will need to send (or else there would be no point in sending them!) But suppose we know that each different message $M_i$ will be sent with probability $p_i$. Can we code the messages $M_i$ into bits in such a way that we can send them through this channel?

The answer is that the total information in the message stream, called the entropy of the stream, is

$$E = \sum_i -p_i \log_2(p_i)$$

bits per message, on average, and the rate at which we can expect to send the messages is no more than $C/E$ messages per second, assuming an optimal translation of messages into bits.

How do you propose to express $E$ without using $\log$?

Here is a third example. Let $\pi(n)$ be the number of prime numbers less than $n$, so for example $\pi(10) = 4$, since 2, 3, 5, and 7 are prime. A famous and deep theorem states that:

$$\pi(n) \sim {n \over \ln n}$$

where $\sim$ means that the ratio of the left and right sides approaches 1 as $n$ becomes very large.

How will you state this without using $\log$?

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I upvoted, but I think the OP may not know what is $\lim$ or $\gamma$ and that a simpler example may suffice – Belgi Aug 28 '12 at 21:03
Wow, not just one special notation for the logarithm. We got two! – Andrea Mori Aug 28 '12 at 21:04
There's $\log_b$ for the generic logarithm in base $b$, and there's $\ln$ for $\log_e$ where $e$ is Euler's number. Unfortunaly MJD edited his answer and $\ln$ disappeared. – Andrea Mori Aug 28 '12 at 21:08
@MJD : well, no problem, actually. Your editing was "unfortunate" only to the extent that made my comment on the two notations quite mysterious. Then also Gustavo edited his comment, making my further explanation unasked for ... :) But it's not really important: we can leave things as they are, no big deal. – Andrea Mori Aug 28 '12 at 21:23
@GustavoBandeira That is not getting rid of the special notation for the logarithm. It is making up your own (confusing, ambiguous) special notation for the logarithm that you use in place of $\log$. – MJD Sep 1 '12 at 18:38

Let's come up with a very simple problem.

Suppose we wanted to write that $\log_2 8 + \log_3 9 = x$ (here, of course, $x = 5$). What would we write without the logarithm notation? We can't write $2^x + 3^x = 8 + 9$, or things along those lines. A priori, we don't know how much of $x$ comes from the $\log_2 8$ term or the $\log_3 9$ term, so we can't write it as two equations.

It's much harder to write this simple equation without logarithms.

As you learn more math, you'll also learn that it can be helpful to take logs of things in general. Taking logs has the effect of "dropping the exponent", i.e. $\log a^b = b \log a$, and the effect of taking products to sums, i.e. $\log(ab) = \log(a) + \log(b)$. Both of these are very nice things to be able to do, as they can simplify hard problems into simpler problems.

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The concept of the logarithm is not just the notation. The logarithm, i.e.a function $\log:\mathbb{R}_{+}\rightarrow\mathbb{R}$, has many usefull properties. The property that made the logarithm so important is that it turns multiplication into addition:


Notice that adding numbers is much simpler than multiplying them. Thus you can multiply various numbers easily if you have a table of logarithms. From a historical perspective that was probably the most important thing that made logarithms so useful and popular among engineers and scientists. So as an important concept it deserves its own symbol: $\log$.

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Having a notation for $\log_2 8$ makes it much easier to express in a succinct way what to do with the solution to the equation after computing it. Writing, say, $$\log_2 x + (\log_2 y)^2$$ is shorter and simpler than writing $$p+q^2 \text{ where }2^p=x\text{ and }2^q=y$$ and it is easier to work with because "$\log_2 y$" tells you everything there is to say about the computation right there in one visual package, whereas "$q \ldots\ldots$ where $2^q=y$" would have your eyes darting back and forth between the use of $q$ and its definition.

The same thing could be said about any intermediate result in a computation that is defined by an equation it satisfies. But defining a new notation each time we meet a new kind of equation would sort of defeat the purpose (they we'd just have to dart back and forth between the definition and its use). What makes cases like logarithms special is that they are useful so often that it is a net timesaver to learn a specialized notation for it once and for all.

There are quite a number of like special cases -- for example a special notation "$a-b$" for subtraction could have been omitted because we can just write it as "the solution to $b+x=a$ instead. And so forth.

Why don't we have a generic notation for "the solution of such-and-such equation" that can be used as an expression? Tradition, mostly. In an ideal world, perhaps we'd write $[x\mathrel{\#}2^x=8]$ instead of $\log_2 8$ and $[x\mathrel{\#}b+x=a]$ instead of $a-b$. But perhaps not. In many cases the traditional notations are for things that are so common that even a generic notation $[x\mathrel{\#}b+x=a]$ would be unacceptably cumbersome compared to $a-b$.

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I wonder if it wouldn't be useful to have a generic notation in addition to the special notations for common cases. Come to think of it, the Hilbert $\varepsilon$ notation is exactly this: $\varepsilon x.2^x=8$ is exactly $\log_2 8$, and $\varepsilon x.b+x = a$ is exactly $a-b$. It might simplify some formulas, but you never see it used. – MJD Sep 1 '12 at 22:24
@MJD: Yes, except that Hilbert's $\varepsilon$ explicitly produces just some value that satisfies the predicate. The use here is probably closer to the "definite description" operator $\iota x.\phi(x)$ going back to Russell (though Russell appears to have written the iota upside-down, if the internet is to be trusted). Of course, that hasn't really caught on either. – Henning Makholm Sep 1 '12 at 23:23

I think if you just want to calculate the logarithm of some number with respect to a certain base, your approach works nicely. But we often deal with the logarithm as a function, so often that this function certainly deserves their own notation.

Addendum: For every positive base and every number, there is one solution of the equation you posted and only one solution. This means, that the equation defines a function, an implicit function. As mentioned in a comment, the squareroot-function is similarly implicitely defined. In calculus, we operate directly with these functions as objects. For example, we may want to calculate the area under the graph of the implicitely defined function. For this, we want to actually talk about the function as an object. Some of these functions, such as the logarithm, occur so often that they deserve their own, special, notation.

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Many texts develop the logarithm function before the exponential function. For example, in M. Spivak's book "Calculus", the function $\log(x)$ is defined for positive $x$ via $\log(x) = \int^{x}_{1}\frac{1}{t}dt.$ Then it follows by the fundamental theorem of calculus that $\log^{\prime}(x) = \frac{1}{x}$ for positive $x.$ Change of variables in integration easily yields the familiar formula $\log(ab) = \log(a) + \log(b)$ for $a,b > 0.$ It then follows that the inverse function to $\log$ is its own derivative, and we obtain the familiar properties of the exponential function. I find his development more natural than, for example, defining $e^{x}$ formally via its power series "out of the blue". Incidentally, it is not so easy to define what you mean by $a^{x}$ for a general base $a$ without using the logarithm at some point.

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What do you need the logarithm for when defining $a^x$? For rational exponents we have $a^{p/q} = \sqrt[q]{a^p}$, and both integral powers and there inverses, integral roots, can be defined without logarithms. In the general case we take a limit. – Yuval Filmus Aug 28 '12 at 22:16
I said it was not so easy- I didn't say it's impossible. Taking that limit in practice is not so straightforward. – Geoff Robinson Aug 29 '12 at 4:55

While pondering the possibility of a less specialized notation, I thought of a specific advantage possessed by the $\log$ notation: it leaves the base of the logarithm implicit. You can say, for example, that $$\log xy = \log x + \log y$$ rather than this version that is more detailed but no more expressive: $$\text{for all positive $a$,}\ \log_a xy = \log_a x + \log_a y$$

Similarly, we can say that the average running time of the quicksort algorithm is $O(n\log n)$, instead of using the unnecessarily specific $O(n\,\log_2 n)$.

And again, but slightly different: the entropy of a message source is $\sum -p_i\log p_i$. Here the logarithm is normally taken to be base 2, but it isn't really important, and taking the logarithms to a different base is just changing the units of measurement, which we normally disregard.

Sometimes the base of the logarithm is important, but often, perhaps most often, it isn't, and it's handy to have a notation that sweeps this unimportant detail under the carpet.

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One reason is that the logarithm is not an elementary function and so cannot be expressed by a "formula".

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Why do they downvote with no reason? =/ – Voyska Aug 28 '12 at 23:36

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