Topic 9 Exponentials and Logarithms

9.1 Introduction

An exponential is something of the form

$y = a^{x}$

E.g.

$y = 2^{x}$

A very common question to ask is something like: if $16 = 2^{x}$ , what is x? In this case the answer is easily seen to be 4. The general case leads us onto logarithms, which are the inverse operation to exponentiation. I.e., if $y = a^{x}$ then $\log_{a} y = x$ — this can be read as “the logarithm to base a of y is x.”

E.g.

$16 = 2^{4}$ means $\log_{2} 16 = 4$
$2^{x} = 32$ means $x = \log_{2} 32$

The most common bases for logarithms are:

10. This is so common that if the base isn’t written, you would normally assume it to be 10, e.g. $\log_{10} 3$ is commonly written as $\log 3$ . This is not, however, a given, so it is worth double checking. The “log” button on most calculators gives logarithms to base 10.
e, where $e = 2.71828...$ is Euler’s number. This is often written as $\ln$ where the “n” stands for “natural.” It is probably the most commonly used base in science and mathematics, and so sometimes $\log x$ might mean $\log_{e} x$ . Most scientific calculators have a “ln” button.

Some modern calculators also have a general $\log_{a} x$ button for general bases.

The definition of a logarithm as the inverse operation to exponentiation

$x = \log_{a} (y) ⟺ a^{x} = y$

gives us the useful results that:

$\begin{aligned} a^{\log_{a} x} & = x \\ \log_{a} (a^{x}) & = x \end{aligned}$

which in particular mean:

$\begin{aligned} 10^{\log_{10} x} & = x \\ \log_{10} (10^{x}) & = x \end{aligned}$

and

$\begin{aligned} e^{\ln x} & = x \\ \ln (e^{x}) & = x \end{aligned}$

9.2 Laws of logarithms

For exponentiation we have the laws of indices:

$\begin{aligned} a^{x} a^{y} & = a^{x + y} \\ \frac{a^{x}}{a^{y}} & = a^{x - y} \\ (a^{x})^{y} & = a^{x y} \end{aligned}$

These then give us the laws of logarithms:

$\begin{aligned} \log_{a} (x y) & = \log_{a} (x) + \log_{a} (y) \\ \log_{a} (\frac{x}{y}) & = \log_{a} (x) - \log_{a} (y) \\ \log_{a} (x^{n}) & = n \log_{a} (x) \end{aligned}$

which, along with the basic definition

$x = \log_{a} (y) ⟺ a^{x} = y$ allow you to manipulate and solve problems with logarithms and exponentiation.

E.g.

$\log_{10} (5 x) = \log_{10} (5) + \log_{10} (x)$
$\log_{3} (x^{2}) = 2 \log_{3} (x)$
$\begin{aligned} \log_{5} (6 x^{2}) & = \log_{5} (6) + \log_{5} (x^{2}) \\ = \log_{5} (6) + 2 \log_{5} (x) \end{aligned}$
$\begin{aligned} \log_{5} (\frac{5}{x^{2}}) & = \log_{5} (5) - \log_{5} (x^{2}) \\ = \log_{5} (5) - 2 \log_{5} (x) \\ = 1 - 2 \log_{5} (x) \end{aligned}$

9.3 Solving equations

Logarithms extend the types of equations that can be solved algebraically.

E.g. $3^{5 x} = 2$ can be solved by taking logarithms of both sides to give: $\begin{aligned} 3^{5 x} & = 2 \\ \log (3^{5 x}) & = \log (2) \\ 5 x \log (3) & = \log (2) \\ x & = \frac{\log (2)}{5 \log (3)} \\ x & = 0.126 \end{aligned}$

Note that this is true whichever base is taken for the logarithm (check), so the base hasn’t been written. The working would have been marginally easier if the base had been taken explicitly to be 3, but not dramatically so.

E.g. $5 \log_{3} x = 4$ can be solved by $\begin{aligned} 5 \log_{3} x & = 4 \\ \log_{3} x & = \frac{4}{5} \\ x & = 3^{\frac{4}{5}} \\ x & = 2.41 \end{aligned}$

E.g. $5 \log_{3} (2 x - 1) = 4$ is very similar

$\begin{aligned} 5 \log_{3} (2 x - 1) & = 4 \\ \log_{3} (2 x - 1) & = \frac{4}{5} \\ 2 x - 1 & = 3^{\frac{4}{5}} \\ 2 x - 1 & = 2.41 \\ x & = \frac{2.41 + 1}{2} \\ x & = 1.70 \end{aligned}$

E.g. $\log_{3} (2 x - 1) + \log_{3} (x - 1) = 1$ just uses the laws of logs first $\begin{aligned} \log_{3} (2 x - 1) + \log_{3} (x - 1) & = 1 \\ (2 x - 1) (x - 1) & = 3^{1} \\ 2 x^{2} - 3 x + 1 & = 3 \\ 2 x^{2} - 3 x + 1 - 3 & = 0 \\ 2 x^{2} - 3 x - 2 & = 0 \\ (x - 2) (2 x + 1) & = 0 \\ x & = 2 or - \frac{1}{2} \end{aligned}$ Which looks great, except you cannot take the logarithm of a negative number so, although both solutions are fine for the first logarithm, taking $x = - \frac{1}{2}$ makes the second one negative, so we only have one solution, $x = 2$ .

LJMU Maths Refresher

Topic 9 Exponentials and Logarithms

9.1 Introduction

9.2 Laws of logarithms

9.3 Solving equations

9.4 Further Resources