Laws of Logarithms for ESAT Mathematics

The Definition of a Logarithm

Logarithms are closely related to indices. They function as the inverse of indices: rather than raising a number to a power to find a result, a logarithm identifies what power a base number must be raised to in order to reach a specific value. This can be expressed by the following general relationship:

$\log_a c = b$ is equivalent to $a^b = c$

To understand this concept, consider some examples using base 10. The expression $\log_{10}$ determines what power 10 must be raised to for a given number:

1. $\log_{10} 10 = 1$ because $10^1 = 10$
2. $\log_{10} 100 = 2$ because $10^2 = 100$
3. $\log_{10} 1000 = 3$ because $10^3 = 1000$
4. $\log_{10} 27 = 1.431363764\dots$ because $10^{1.431363764\dots} = 27$

Logarithms can use other bases as well. For base 2, we find:

1. $\log_2 32 = 5$ because $2^5 = 32$
2. $\log_2 \frac{1}{2} = -1$ because $2^{-1} = \frac{1}{2}$

Important Constraints

There are three critical rules regarding the values used in logarithms that you must remember for the ESAT:

1. The base number $a$ must be positive ($a > 0$) and cannot be equal to 1 ($a \neq 1$).
2. We can only take the logarithm of positive numbers, meaning $c > 0$. The log function is not defined for zero or negative numbers.
3. The result of a logarithm, $b$, can be any number, including negative values or zero.

Exercises in Definition

You should be able to evaluate simple logarithmic expressions by inspection:

1. $\log_5 5^2 = 2$
2. $\log_3 3^5 = 5$
3. $\log_7 \sqrt{7} = \log_7 7^{1/2} = 0.5$

Graphical Representation of Logarithms

We can understand logarithms graphically by comparing them to exponential functions. Consider $y = 2^x$. This function maps values from the $x$-axis to the $y$-axis.

If we start on the $y$-axis (for example at 8) and trace back to the corresponding value on the $x$-axis (which is 3), we are performing a logarithmic operation: $y \to \log_2 y$. The graph of $y = \log_2 x$ is simply the graph of $y = 2^x$ with the $x$ and $y$ axes swapped.

Note that the graph is only defined for $x > 0$. It crosses the $x$-axis at 1 because $2^0 = 1$, which means $\log_2 1 = 0$.

The Laws of Logarithms

You must be able to use the following rules to manipulate and simplify expressions. These laws are direct equivalents of index laws.

The Addition Law

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

This is the logarithmic equivalent of $a^p a^q = a^{p+q}$. We can derive it by noting that $a^{\log_a x + \log_a y} = a^{\log_a x} a^{\log_a y} = xy = a^{\log_a(xy)}$. This relies on the identity $a^{\log_a x} = x$, which follows from the definition of a logarithm.

The Subtraction Law

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

Derivation: $a^{\log_a x - \log_a y} = \frac{a^{\log_a x}}{a^{\log_a y}} = \frac{x}{y} = a^{\log_a (x/y)}$.

The Power Law

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

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

Special Cases

1. $\log_a \left(\frac{1}{x}\right) = -\log_a x$: This is a specific case of the power law where $k = -1$.
2. $\log_a a = 1$: This is true because $a^1 = a$.

The Change of Base Formula

Although questions requiring this formula will not be set in the ESAT, it is a valuable tool for your mathematical toolkit. It allows you to convert a logarithm from one base to another:

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

For example, to find $\log_4 23$ in terms of $\log_7$, we set $p = \log_4 23$, which means $4^p = 23$. Taking $\log_7$ of both sides gives $\log_7 4^p = \log_7 23$, then $p \log_7 4 = \log_7 23$. Thus, $p = \frac{\log_7 23}{\log_7 4}$.

A useful extension of this occurs when $c = b$, leading to $\log_a b = \frac{1}{\log_b a}$.

Solving Exponential Equations

You can use logarithms to solve equations of the form $a^x = b$. Often, you will be required to give an exact answer rather than a decimal approximation.

Example: Solve $5^{2x} = 27$ exactly.

Approach 1: Use base 5 Take $\log_5$ of both sides: $\log_5 5^{2x} = \log_5 27$. Since $\log_5 5^{2x} = 2x$ and $27 = 3^3$, we have $2x = \log_5 3^3$. Using the power law: $2x = 3 \log_5 3$. Therefore, $x = \frac{3}{2} \log_5 3$.

Approach 2: Use base 3 Take $\log_3$ of both sides: $\log_3 5^{2x} = \log_3 27$. Using the power law: $2x \log_3 5 = \log_3 3^3$. Since $\log_3 3^3 = 3$, we have $2x \log_3 5 = 3$. Therefore, $x = \frac{3}{2 \log_3 5}$.