What is the logarithm?

When students first learn about the logarithm, they ask “What IS the logarithm?”

Before answering that, let’s start with something that you already understand, and then I will connect it to the logarithm.

Inverses

Let’s start with the concept of the inverse. And let’s start with simple, intuitive examples.

What is the value of \(5+3-3\)? Of course the plus and minus 3 cancel, so the value is 5.

What is the value of \(5\cdot 3/3\)? Of course the times and divide 3 cancel, so the value is 5.

Plus and minus are inverses of each other, in some sense—and times and divide are inverses of each other.

This is important for solving equations. If you have the equation \(x+3 = 10\) then to solve it we apply the inverse:

$$ x+3 - 3 = 10-3 $$

which gives us \(x=7\).

Or if we want to solve \(3x = 10\) again we use the inverse:

$$ 3x/3 = 10/3 $$

and therefore \(x=10/3\).

Inverse of Exponents

We’ve seen the inverses of addition and multiplication.

What about exponents? What is the inverse of an exponent?

One version of this is already familiar to you. \( \sqrt{ 5^2 } = 5\). The square-root and the square cancel each other, so they are inverses. It doesn’t matter if we write it this way or \( (\sqrt 5)^2 =5\).

However, be careful! This only works with positive numbers. For negative numbers, the square-root and square are not perfect inverses! Observe that \( \sqrt{(-5)^2} = 5\).

And \( (\sqrt {-5})^2\) isn’t even defined, if we are only using real numbers. That is because \( \sqrt{-5}\) isn’t a real number. (It is a “complex number” but let’s leave that topic for another day.)

So keep in mind that the square-root and square are inverses only if we deal with nonnegative numbers. (“Non-negative” means “positive or zero”. The square-root and square are inverses for all positive numbers and for zero.)

This is useful in solving equations. In the equation \( x^2 = 100 \) we apply the inverse. But keep in mind that this only works for non-negative x.

For non-negative x, we have

$$ \sqrt{x^2} = \sqrt{100} $$

and therefore \( x = 10\).

If you want the solutions for all x then you must include the negatives. If x is negative then

$$ -\sqrt {x^2} = \sqrt {100} $$

and therefore \(-x = 10\) which implies \(x = -10\).

Or if we want to solve \( \sqrt x = 100 \) then

$$ (\sqrt x)^2 = 100^2 $$

and therefore \(x = 10000\).

For this equation, there are no negative solutions for x.

Of course this all uses the square-root and the square, but it generalizes to other exponents. The inverse of the cube is the cube-root, and the inverse of the fourth exponent is the fourth root—again, for nonnegative x. For negative numbers it all depends on whether the exponent is odd or even.

I can talk about all of this more in a different post if anyone requests it, but we’re trying to talk about the logarithm, so let’s move along to the logarithm.

Inverses of variable exponents

Ok, we have seen that the inverse of the nth exponent is the nth root, for positive x.

But that is for a variable base and a specific exponent. What if we did it the other way around?

What if the variable is in the exponent and the base is fixed? For example, consider \(2^x\). What is the inverse of this?

That is the logarithm!

The logarithm is the inverse of the variable exponent!

If the base of the exponent is 2, like in \(2^x\), then its inverse is “log base 2” which we write as \(\log_2 x\).

So the inverse of \(2^x\) is \(\log_2 x\).

Remember, the idea of an inverse is that it causes a cancellation. Therefore \(\log_2(2^x) = x\) in the same way that \(x+3-3 = x\) and \(x\cdot 3/3 = x\) and for positive numbers, \(\sqrt{ x^2} = x\).

Of course this works for any base, not just base 2. We also have \(\log_3(3^x) = x\) and \( \log_\pi(\pi^x) = x\), and so on.

… So there you have it! You now know what the logarithm is!

But wait … how do you … calculate with it?

Ok, you know what the logarithm is. But that might not make it obvious, how you calculate \(\log_2(16)\).

How do you actually find logarithm values?

Well, you use the cancellation property! This requires taking the number inside and writing it as an exponent, so that then the exponent and logarithm will cancel and simplify.

For example, we know that \(16 = 2^4\). Therefore

$$ \begin{aligned} \log _ 2 (16) &= \log _ 2 (2^4) \\ &=4 \end{aligned} $$

So there you have it! The logarithm calculated!

Let’s do another. Let’s find \(\log _ 3 (1)\). Again we have to express 1, this time as an exponent of 3. Well we know that \(3^0 = 1\). Therefore

$$\begin{aligned} \log _ 3(1) &= \log _ 3(3^0) \\ &= 0 \end{aligned} $$

So again we are able to find the logarithm if we can express the number as an exponent.

Let’s see another, let’s find \(\log_4(1/4)\). Again we want to express 1/4 as an exponent of 4. We know that \(4^{-1}=1/4\). Therefore

$$\begin{aligned} \log _ 4(1/4) &= \log _ 4 (4^{-1}) \\ &= -1 \end{aligned}$$

Now you might wonder “What about \(\log_2(3)\)? What if you can’t find 3 as an exponent of 2?”

Well sure, that’s much more complicated and you typically won’t be asked to do that by hand. You can put it in a calculator, or leave the answer as just \(\log_2(3)\).

And there are certain other logarithm calculations that are trickier, like \(\log_{1/4}(16)\), or things that involve “change of base”. There are multiplication and exponent laws for logarithms. So the topic gets more complex from here.

But at least for a fundamental understanding of what the logarithm is, you now have it!