Pendahuluan logaritma

Deriving and Understanding Log Rules Using the Definition

Logarithms have some pretty cool properties. Let's dive in and check some of them out.

Recall (or maybe not: that's okay!) the definition of a logarithm. A logarithm $L = \log_b x$ is the solution to the equation $b^L = x$. From now on, I'll leave out the base when I'm writing logarithms. You can assume that I mean base $e$.

First, a trivial one: $e^{\log a} = a$. If you don't see why this is true, think about it for a bit.

It follows from the definition of $\log a$. Let's look at some more interesting stuff :)

The second identity we'll tackle is that $\log a + \log b = \log ab$.

We don't really know much about what logarithms are, yet, so what should we do next? Think about it for a sec.

We'll write out the definition of $\log a$ and $\log b$, and good things should happen!

We have $$\begin{aligned} e^{\log a} &= a, \\ e^{\log b} &= b. \end{aligned}$$ Multiplying the equations according to our intuition gives $e^{\log a} \cdot e^{\log b} = e^{\log a + \log b} = ab.$ To finish, let's go back to logarithms by taking the $\log$ of both sides to get $\log a + \log b = \log ab$, as desired. $\square$

Note that $e$ could be replaced with any other letter, so of course this proof holds for any base (if it helps, think of $e$ as a variable rather than a number to convince yourself).

If we got something this nice by adding logs, surely, there exists something similar by subtracting logs, right? Try working it out.

We use the exact same procedure as before, except we divide $a = e^{\log a}$ by $b = e^{\log b}$, and we get $\log a - \log b = \log (a/b)$.

Next we'll prove that $\log a^k = k\log a$. An implication of this identity is that $\log \sqrt[n]{a} = \frac{\log a}{n}$.

We're going to use $(e^x)^y = e^{xy}$; we'll start by writing the defintion of $\log a^k$ and then make some manipulations. $$\begin{aligned} e^{\log a^k} &= a^k \\ \left(e^{\log a^k}\right)^{\log a} &= \left(a^k\right)^{\log a} \\ \left(e^{\log a}\right)^{\log a^k} &= a^{k\log a} \\ a^{\log a^k} &= a^{k\log a} \\ \log_a \left(a^{\log a^k}\right) &= \log_a \left(a^{k\log a}\right) \\ \log a^k &= k\log a. \square \end{aligned}$$

Though this proof is nice and straightforward, I have another proof. I like it a lot, but unfortunately it only works when $k$ is a positive integer.

We want to prove that $k\log a = \log a^k$:

$$k\log a = \underbrace{\log a + \log a + \cdots + \log a}_{k \ \log a\text{'s}} = \log(\underbrace{a\cdot a \cdots a}_{k \ a\text{'s}}) = \log a^k.$$

The above proof is another reason why this property should be fairly intuitive to you.

Finally, I'll show a quick proof of the change-of-base formula. I don't really care too much for this one since I rarely run into base issues, but it's useful to know for problems that intentionally force base change.

The formula states that $\log_m n = \frac{\log_b n}{\log_b m}$. (An editor will have to help me with some intuition for this one!)

Let $\phi = \log_m n$. Then $m^\phi = n$, so $$\log_b n = \log_b (m^\phi) = \phi \log_b m,$$ and substituting for $\phi$ gives $\log_b n = (\log_m n) (\log_m b)$ which implies $\log_m n = \frac{\log_b n}{\log_b m}$, as desired. $\square$

Well, that about covers most useful logarithm identities. Don't worry too much about memorizing these properties; hopefully, seeing the proofs have made them intuitive for you, so that you'll pull them out your sleeve as necessary!

What is a logarithm?

A logarithm is the inverse function to exponents. If you don't have a full grasp on exponents, please click here to refresh yourself!

We will begin with looking at logarithms to base $10$. There is nothing special about the number $10$, we just use it because we use a decimal system of counting, which are in tenths, hundredths, thousandths, etc.

We will consider the function $y = 10 ^x$. Given any number $x$, we can obtain another number which will be equal to $y$ in the equation above.

For example, let's say that we have $13 = 10 ^ x$. We can think of this as wanting to find the power to which $10$ must be raised to obtain $13$. So, in order to solve for $x$, we must take the log of both sides, which will look like: $$\log_{10} 13 = x$$

We can easily punch the left side into our calculator, to give us $x = 1.11394225231$. This means that $13$ is equal to $10^{1.11394225231}$.

Graphs of $\log_{10}{(x)}$

The above graph is the visual depiction of $y = 10 ^ x$. As you can see, it is never negative and never takes the value $0$. There is no value to raise $10$ to obtain a number less than or equal to $0$. We say that $\log_{10}{(x)}$ is undefined for $x \leq 0$. Now, let's look at the graph of $y =\log_{10}{(x)}$

The graph does not stay below a certain height as $x$ gets large. There is no horizontal asymptote. This function increases very slowly as $x$ increases.