I have heard of logarithms, and done very little research at all. From that little bit of research I found out its in algebra 2. Sadly to say, I'm going into 9th grade, but yet I'm learning [calculus!?] and I don't know what a logarithm is! I find it in many places now. I deem it important to know what a logarithm is even though I'm jumping the gun in a sense. My understanding of concepts, is just like that of programming. In the mean time, you know its there, and your ITCHING SO HARD to find out what that is, but nope! For now we use it, tomorrow we learn what it does.

I just know that to identify a logarithm at my level, I just look for a log. :P

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    $\begingroup$ wikipedia. $\endgroup$ – Karolis Juodelė Mar 22 '14 at 13:27
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    $\begingroup$ It is the inverse operation to exponentiation. Careful, not of powering. While $\sqrt{e^2}=e$, $\log(e^2)=2$. $\endgroup$ – geodude Mar 22 '14 at 13:27
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    $\begingroup$ How do you explain the concept of logarithm to a five year old? :-) $\endgroup$ – Raymond Manzoni Mar 22 '14 at 13:33
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    $\begingroup$ @LinkTheProgrammer that's a little immature don't you think? Wikipedia is a very reliable source of information. $\endgroup$ – Guy Mar 22 '14 at 13:42
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    $\begingroup$ In the end, $1 + 1$ stays as $2$, even on Wikipedia. It should be credible enough for mathematics related information. $\endgroup$ – Yiyuan Lee Mar 22 '14 at 13:52

If you know what a power function is:


you can choose to solve for $a$ or $b$. If you want $a$, take $b$-th root on both sides:


Imagine $b=2$.

However, if you want to get $b$, you take the logarithm:


Here, I used the logarithm with a base $a$. Logarithms of different bases are related: they are simple multiples of each other. Common logarithms are $\log_{10}$ (the base is usually skipped), and the natural logarithm ($\log_e x=\ln x$) which is a very nicely-behaved function when you go further in calculus.

The numerical meaning of logarithm can be roughly understood as this: the whole part of the value of $\log_{10} x$ counts the number of digits in $x$. For instance

$$\log 1=0$$ $$\log 10=1$$ $$\log 100=2$$ $$\log 1000=3$$

and so on. Of course, you can evaluate something like $$\log 500=2.69...$$ $$\log 0.05=-1.30...$$

The rest of the properties follow from the definition that it inverts $a^b=c$. For instance, logarithm of a product can be split into sum of logarithms:

$$\log{ab}=\log a + \log b$$

Ultimately, it's just another elementary function, like roots, polynomials and so on.

  • a ++ $=$ ++ a, so incrementation has a single inverse operation, namely decrementation.

  • Repeated incrementation is addition.

  • $a+b=b+a$, so addition has a single inverse operation, namely subtraction.

  • Repeated addition is multiplication.

  • $a\cdot b=b\cdot a$, so multiplication has a single inverse operation, namely division.

  • Repeated multiplication is exponentiation.

  • But $a^b\neq b^a$ $\big($usually$\big)$, so exponentiation has not just one, but two inverse operations: root extraction $\big($when the exponent is known, and we want to find out the value of the base$\big)$, and logarithms $\big($for the opposite case: namely when the base is known, and we want to find out the value of the exponent$\big)$.

  • $\begingroup$ It beats me what the first line could possibly mean mathematically. In programming (C like languages) it is undefined behaviour. There also seems to be a consistent suggestion that there is a link between commutativity and the existence of inverses; this is wrong as all elements in non-Abelian groups have inverses, but those in commutative monoids do not. $\endgroup$ – Marc van Leeuwen Mar 22 '14 at 19:58
  • $\begingroup$ @Marc It's defined behavior in Java and other languages, but will be false in all cases (although no sane person has ever written such a statement except to test the previous statement). And yes this seems pretty confusing to me too. $\endgroup$ – Voo Mar 22 '14 at 20:11
  • $\begingroup$ Just out of curiosity: is the a++ syntax familiar for non-programmers? $\endgroup$ – streppel Mar 22 '14 at 22:07
  • $\begingroup$ @Streppel no, its not. But he realizes that using an analogy may help me, which it does because I program in Java. However the answer above his helps to explain it. :D $\endgroup$ – AMDG Apr 25 '14 at 23:29

Logarithms are the inverse to exponentiation. Simply put, $\log_a(k)$ is the solution to the equation


Here $a$ is called the "base" of the logarithm.

Obviously, $k\gt0$ and therefore, the logarithms of negative numbers are not defined, as long as we are dealing with the set of real numbers. Also $1^x=k$ will have no solutions for $k\ne1$, therefore $a$ must be not equal to $1$.

$(-1)^x$(or any other negative base) isn't well defined for non-integral $x$ and therefore $a\gt0$.

This sets some constraints on the domain of $\log$

  • $a\ne1$
  • $k\gt0$
  • $a\gt0$

$\log_e(x)$ is called the natural logarithm, sometimes abbreviated as $\ln(x)$ or simply $\log(x)$


The function whose values at $x$ are $c\log{x}$ where $c$ is a constant is the most general function that satisfies, for all positive $x$ and $y$, the identity


Let $x\to e^x$ and $y\to e^y$, then we have $h(e^x)+h(e^y)=h(e^xe^y)=h(e^{x+y})$

Let $H(u)=h(e^u)$, then we have $H(x)+H(y)=H(x+y)$.

Let $x=y$, then we have $2H(x)=H(2x)\implies H(x)=\frac{1}{2}H'(2x)$.

Differentiating gives $H'(x)=H'(2x)$.

Repeated differentiation gives the general formula, $H^{(n)}(x)=2^{n-1}H^{(n)}(2x)$ for $n\ge0$.

Let $x=0$, then we have $(2^{n-1}-1)H^{(n)}(0)=0$

So $H^{(n)}(0)=0$ for all integers $n\ne 1$.

By considering the Maclaurin series for $f(x)$, we find that the most general function satisfying $H(x)+H(y)=H(x+y)$ is $cx$ where $c$ is a constant.

So $h(e^x)=cx$. Letting $x\to \log{x}$ we find the most general function satisfying the original function equation is $c\log{x}$.

  • $\begingroup$ Are you a programmer? you seem to have efficiently ordered the steps quite nicely :) $\endgroup$ – AMDG Apr 25 '14 at 23:32

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