Laws of Logarithms

Laws of logarithms

$a^b = c \Leftrightarrow b = \log_a c$

$\log_a x + \log_a y = \log_a(xy)$

$\log_a x - \log_a y = \log_a \left(\frac{x}{y}\right)$

$k \log_a x = \log_a(x^k)$

including the special cases:

$\log_a \frac{1}{x} = -\log_a x$

$\log_a a = 1$

Questions requiring knowledge of the change of base formula will not be set.

Logarithms are very closely related to indices; in fact, they are really the "inverse of indices". They tell you what power a number has to be raised to rather than raising a number to a power. Let's unpack that idea a little bit to get an idea of how logs work. We will start with a few examples to help you get the feel for things and then we will look at logarithms graphically; and then we will move on to exploring [briefly!] how the logarithm rules work.

But before we begin, a teeny little bit of history is useful. Before calculators existed, doing lots of calculations, especially with big numbers, could be complicated. So, logarithms were invented to make the calculations easier [it is worth looking at the history a little bit to understand how clever and inventive mathematicians can be – see the Wikipedia page on the History of Logarithms. Even if you are not keen on history, the background to logarithms is worth exploring.]

Here are some examples using $\log_{10}$. $\log_{10}$ tells you what power 10 needs to be raised to get a given number:

• $\log_{10} 10 = 1$ because 10 needs to be raised to the power of 1 to get 10: $10^1 = 10$
• $\log_{10} 100 = 2$ because 10 needs to be raised to the power of 2 to get 100: $10^2 = 100$
• $\log_{10} 1000 = 3$ because 10 needs to be raised to the power of 3 to get 1000: $10^3 = 1000$
• $\log_{10} 27 = 1.431363764 \dots \dots$ because $10^{1.431363764\dots} = 27$ [you can check this with your calculator]

We can also try using logs to other "bases". $\log_2$ tells you what power 2 needs to be raised to get a given number: here are some other [reasonably obvious] examples:

• $\log_2 32 = 5$ because 2 needs to be raised to the power of 5 to get 32: $2^5 = 32$
• $\log_2 \frac{1}{2} = -1$ because 2 needs to be raised to the power of $-1$ to get $\frac{1}{2}$: $2^{-1} = \frac{1}{2}$

So, we have in general the following relationship between log and powers:

$\log_a c = b \quad \text{says the same as} \quad a^b = c$

You should make sure you are very familiar and comfortable with this idea.

A few things to note about this [things might change when you meet more mathematics but then definitions get honed and changed too]:

• We only take logs with a positive base number: so $a > 0$ [but $a \neq 1$]
• We can only take the logs of positive numbers: so $c > 0$
• The log of a number can be negative: so $b$ can be any number [even 0]

You should be able to work out why all three of these statements apply – look back to our discussion on indices if you are not sure. It is important to remember that the log function is not defined for negative numbers – that is we must have $c > 0$. [This was important in a TMUA/ESAT question from a few years ago: TMUA paper 1 2021 question 20.]

Exercise

Work out each of the following:

$\log_5 5^2 ; \log_3 3^5 ; \log_7 \sqrt{7}$

We can take a brief look at logs and graphs. We will work with base 2 as that gives nice graphs.

First, we draw $y = 2^x$ and look at a few values:

From this, you can see that $2^x$ takes numbers from the $x$-axis and gives us numbers on the $y$-axis: it maps $x \to 2^x$. And also, if you start on the $y$-axis [say with 8] and trace back to what number corresponds to it on the $x$-axis you get the log of the number of the $y$-axis, namely 3. So going backwards from the $y$-axis to the $x$-axis we get $y \to \log_2 y$.

We can now draw the log graph as it is just the graph of $y = 2^x$ with the $x$ and $y$ axes swapped [use a graph package to draw $y = 2^x$ and $y = 10^x$ to see how fast they grow – that is exponential growth]:

A few things to note here too:

• You can see the graph is only defined for $x > 0$ as we expected.
• The graph of $y = \log_2 x$ crosses the $x$-axis at 1. Can you explain why? [Because $2^0 = 1$ and so $\log_2 1 = 0$.]

Now we can look at all the logarithm rules we expect you to know – you should make sure you understand them [i.e. you know how they work and where they come from] and you should make sure you can use them correctly.

We start with $\log_a x + \log_a y = \log_a(xy)$. This is really the logarithm equivalent of $a^p a^q = a^{p+q}$ [think about how this relates to the log equation]. We can see how this equation works as follows:

$a^{\log_a x + \log_a y} = a^{\log_a x} a^{\log_a y} = xy = a^{\log_a(xy)}$

And make sure you can see what rule we have used at each stage of this. Note we have used one idea that we haven't drawn attention to as yet: $a^{\log_a x} = x$. We hope this idea is "obvious" as it is essentially the very definition of a logarithm.

We can now look at the other rules in the same way:

$\log_a x - \log_a y = \log_a \left(\frac{x}{y}\right)$

$a^{\log_a x - \log_a y} = a^{\log_a x} a^{-\log_a y} = \frac{a^{\log_a x}}{a^{\log_a y}} = \frac{x}{y} = a^{\log_a \left(\frac{x}{y}\right)}$

$k \log_a x = \log_a (x^k)$

$a^{k \log_a x} = (a^{\log_a x})^k = x^k = a^{\log_a (x^k)}$

$\log_a \frac{1}{x} = -\log_a x$

$a^{-\log_a x} = \frac{1}{a^{\log_a x}} = \frac{1}{x} = a^{\log_a \frac{1}{x}}$

And finally

$\log_a a = 1$

which should be "obvious" because $a^1 = a$.

The specification mentions the change of base formula and says it will not be examined in TMUA/ESAT. Nevertheless, it is a useful formula and we recommend you have it in your "remembered formulae" maths kit, and make sure you can derive it and understand it too! We will take a brief look at the formula here [but you can skip this section as it is not part of the TMUA/ESAT and we won't ask questions that depend on it].

The change of base formula allows you to convert from a log with one base to a log with another base; for instance, changing from base 4 to base 7: $\log_4$ to $\log_7$.

Before we explore this idea, have a think about how you might go about this task; for instance, how might you find $\log_4 23$ in terms of $\log_7 23$?

Let's start with the example we just gave:

Example

Find $\log_4 23$ in terms of $\log_7 23$.

Let $\log_4 23 = p$, then $4^p = 23$. Take $\log_7$ of both sides: $\log_7 4^p = \log_7 23$, which gives $p \log_7 4 = \log_7 23$, so

$p = \log_4 23 = \frac{\log_7 23}{\log_7 4}$

We can use the same method to derive the change of base formula:

Example

Change from $\log_a b$ to $\log_c b$.

Let $p = \log_a b$, then $a^p = b$. Take $\log_c$ of both sides: $\log_c a^p = \log_c b$, which gives $p \log_c a = \log_c b$, so

$p = \log_a b = \frac{\log_c b}{\log_c a}$

and this gives us the change of base formula:

$\log_a b = \frac{\log_c b}{\log_c a}$

And as a final note, this leads to one more useful formula if we set $c = b$:

$\log_a b = \frac{\log_b b}{\log_b a}$

and as $\log_b b = 1$ we have:

$\log_a b = \frac{1}{\log_b a}$