Logarithmic and Exponential Equations (Senior 5 Mathematics): free lessons | Ganzaa

Logarithmic and Exponential Equations

Lessons

Senior 5 · Mathematics

Logarithmic and Exponential Equations

4 lessons

Lesson 1 of 4

Exponential and logarithmic functions

25 min

Which power of 2 is 1000?

$2^{9} = 512$ and $2^{10} = 1024$ , so the power of 2 that gives 1000 lies just below 10. In S4 you used decimal logarithms to answer questions like this for powers of 10. This unit works with any positive base: the exponential function $a^{x}$ , which models growth and decay, and its inverse, the logarithm $\log_a x$ , which finds the exponent. The answer to the question above is $\log_2 1000 \approx 9.97$ , and by the end of this lesson you will be able to compute it with an ordinary calculator.

Definition

Exponential function with base $a$ ( $a \gt 0$ , $a \ne 1$ ): the function $f (x) = a^{x}$ , defined for every real $x$ .

Example: $f (x) = 3^{x}$ gives $f (2) = 9$ , $f(-1) = \frac{1}{3}$ and $f\left(\frac{1}{2}\right) = \sqrt{3}$ .

Graphs and properties of $a^{x}$

Figure 1.

y = 2^{x}

(increasing, base

\gt 1

) and

y = \left(\frac{1}{2}\right)^x = 2^{-x}

(decreasing, base between 0 and 1). Both pass through

(0, 1)

and have the

x

-axis as asymptote.

Property of $y = a^{x}$	$a \gt 1$	$0 \lt a \lt 1$
Domain and range	domain $\mathbb{R}$ , range $(0, +\infty)$ : $a^{x}$ is never 0 or negative
Passes through	$(0, 1)$ , since $a^{0} = 1$ , and $(1, a)$
Variation	strictly increasing	strictly decreasing
Asymptote	$y = 0$ as $x \to -\infty$	$y = 0$ as $x \to +\infty$

The laws of indices from S1–S4 hold for real exponents: $a^x a^y = a^{x+y}$ , $\frac{a^x}{a^y} = a^{x-y}$ , $(a^x)^y = a^{xy}$ and $a^{-x} = \frac{1}{a^x}$ . Because $a^{x}$ is strictly monotone, it never takes the same value twice: $a^{x} = a^{y}$ implies $x = y$ . Lesson 2 uses this fact to solve equations.

Definition

Logarithm to base $a$ ( $a \gt 0$ , $a \ne 1$ ) of a positive number $x$ : the exponent $y$ to which $a$ must be raised to give $x$ .

y = \log_a x \iff a^y = x \qquad (x \gt 0).

Notation: $\log x$ (or $\lg x$ ) means $\log_{10} x$ , the common (decimal) logarithm of S4. $\ln x$ means $\log_e x$ , the natural logarithm, whose base is the number $e \approx 2.718$ ; it has its own calculator key, and you will study $e$ in S6.

Example: $\log_2 8 = 3$ because $2^{3} = 8$ ; $\log_{10} 0.001 = -3$ because $10^{-3} = 0.001$ .

The logarithm is the inverse of the exponential

The function $x \mapsto \log_a x$ undoes $x \mapsto a^x$ : $\log_a(a^x) = x$ for every real $x$ , and $a^{\log_a x} = x$ for every $x \gt 0$ . So its graph is the reflection of the graph of $a^{x}$ in the line $y = x$ , its domain is $(0, +\infty)$ and its range is $\mathbb{R}$ . It passes through $(1, 0)$ and $(a, 1)$ : $\log_a 1 = 0$ and $\log_a a = 1$ . For $a \gt 1$ it is increasing, and the $y$ -axis is a vertical asymptote.

Figure 2.

y = \log_2 x

is the reflection of

y = 2^{x}

in the line

y = x

: swapping the coordinates of every point turns one graph into the other.

Worked example 1: evaluating logarithms

Find (a) $\log_2 32$ (b) $\log_9 27$ (c) $\log_{\frac{1}{3}} 81$ (d) $\log_5 \frac{1}{125}$ .

Solution In each case, write the number as a power of the base.

$32 = 2^{5}$ , so $\log_2 32 = 5$ .
$9^{y} = 27$ means $3^{2y} = 3^3$ , so $2 y = 3$ and $\log_9 27 = \frac{3}{2}$ .
$\left(\frac{1}{3}\right)^y = 81$ means $3^{-y} = 3^4$ , so $\log_{\frac{1}{3}} 81 = -4$ .
$\frac{1}{125} = 5^{-3}$ , so $\log_5 \frac{1}{125} = -3$ .

Important

Laws of logarithms ( $a \gt 0$ , $a \ne 1$ , $x \gt 0$ , $y \gt 0$ , $k$ real)

\log_a(xy) = \log_a x + \log_a y \qquad \log_a\frac{x}{y} = \log_a x - \log_a y \qquad \log_a x^k = k\log_a x

\log_a 1 = 0, \qquad \log_a a = 1, \qquad \log_a\frac{1}{x} = -\log_a x.

Why the first law holds: put $p = \log_a x$ and $q = \log_a y$ . Then $x = a^{p}$ and $y = a^{q}$ , so $xy = a^{p+q}$ , which says $\log_a(xy) = p + q$ . The others follow in the same way from the laws of indices.

Change of base: for any other base $b$ ,

\log_a x = \frac{\log_b x}{\log_b a}.

Why: if $y = \log_a x$ then $a^{y} = x$ ; take $\log_b$ of both sides: $y\log_b a = \log_b x$ . With $b = 10$ or $b = e$ , a calculator gives any logarithm.

Worked example 2: using the laws

(a) Simplify $2\log 5 + \log 4$ . (b) Simplify $\log_3 54 - \log_3 2$ . (c) Express $\log_a\frac{x^3\sqrt{y}}{z^2}$ in terms of $\log_a x$ , $\log_a y$ and $\log_a z$ .

Solution

$2\log 5 + \log 4 = \log 25 + \log 4 = \log 100 = 2$ .
$\log_3 54 - \log_3 2 = \log_3 27 = 3$ .
$\log_a x^3 + \log_a y^{\frac{1}{2}} - \log_a z^2 = 3\log_a x + \frac{1}{2}\log_a y - 2\log_a z$ .

Worked example 3: change of base and domains

(a) Find $\log_3 20$ and $\log_2 1000$ to 3 decimal places. (b) Find the domain of $f(x) = \log_2(x^2 - 4)$ and of $g(x) = \log_3(5 - x) + \log_3(x + 1)$ .

Solution

$\log_3 20 = \frac{\log 20}{\log 3} = \frac{1.30103}{0.47712} = 2.727$ , and $\log_2 1000 = \frac{\log 1000}{\log 2} = \frac{3}{0.30103} = 9.966$ .
A logarithm needs a positive argument. For $f$ : $x^2 - 4 \gt 0$ , so $x \lt -2$ or $x \gt 2$ : the domain is $(-\infty, -2) \cup (2, +\infty)$ . For $g$ : $5 - x \gt 0$ and $x + 1 \gt 0$ , so the domain is $(- 1, 5)$ .

Warning

Three false "laws". $\log_a(x + y) \ne \log_a x + \log_a y$ (try $\log 5 + \log 5 = \log 25$ , not $\log 10$ ); $\frac{\log_a x}{\log_a y} \ne \log_a\frac{x}{y}$ ; and $(\log x)^2 \ne \log x^2 = 2\log x$ . Also, $\log_a 0$ and $\log_a(\text{negative})$ do not exist: no power of a positive base is 0 or negative.

Tip

A logarithm is an exponent. Read $\log_a x$ as "the power of $a$ that gives $x$ ". Every law of logarithms is a law of indices in disguise: multiplying powers adds exponents, so the log of a product is a sum.

Khan Academy: Introduction to Logarithms (about 7 min). Watch how Sal turns the question "2 to what power gives 16?" into the logarithm, and then evaluates logarithms with other bases.

✦

Practice

The last two are harder.

Write in logarithmic form: (a) $3^{4} = 81$ (b) $10^{-2} = 0.01$ . Write in exponential form: (c) $\log_5 125 = 3$ (d) $\log_4 2 = \frac{1}{2}$ .
Evaluate (a) $\log_2 64$ (b) $\log_{27} 3$ (c) $\log_4\frac{1}{8}$ (d) $\log_{\sqrt{2}} 8$ .
Simplify (a) $\log 8 + \log 125$ (b) $\log_2 40 - \log_2 5$ (c) $\frac{1}{2}\log_3 81$ .
Use the change of base to find $\log_7 50$ to 3 decimal places.
Given $\log_a 2 = p$ and $\log_a 3 = q$ , express (a) $\log_a 12$ (b) $\log_a\frac{9}{8}$ (c) $\log_a\sqrt{6}$ in terms of $p$ and $q$ .
Find the domain of (a) $\log(x - 3)$ (b) $\log_2(9 - x^2)$ .
Sketch $y = 3^{x}$ and $y = \log_3 x$ on the same axes. Give the domain, the range and the axis intercept of each.
Prove that $\log_a b \times \log_b c \times \log_c a = 1$ for positive $a, b, c \ne 1$ .

Show answersHide answers

(a) $\log_3 81 = 4$ (b) $\log 0.01 = -2$ (c) $5^{3} = 125$ (d) $4^{\frac{1}{2}} = 2$ .
(a) 6 (b) $\frac{1}{3}$ (c) $2^{2y} = 2^{-3}$ , so $-\frac{3}{2}$ (d) $2^{\frac{y}{2}} = 2^3$ , so 6.
(a) $\log 1000 = 3$ (b) $\log_2 8 = 3$ (c) $\log_3 9 = 2$ .
$\frac{\log 50}{\log 7} = \frac{1.69897}{0.84510} = 2.010$ .
(a) $\log_a(2^2 \times 3) = 2p + q$ (b) $2 q - 3 p$ (c) $\frac{1}{2}(p + q)$ .
(a) $x \gt 3$ (b) $9 - x^2 \gt 0$ : $-3 \lt x \lt 3$ .
$y = 3^{x}$ : domain $\mathbb{R}$ , range $(0, +\infty)$ , meets the $y$ -axis at $(0, 1)$ . $y = \log_3 x$ : domain $(0, +\infty)$ , range $\mathbb{R}$ , meets the $x$ -axis at $(1, 0)$ . They are reflections of each other in $y = x$ .
By the change of base to base 10: $\frac{\log b}{\log a} \times \frac{\log c}{\log b} \times \frac{\log a}{\log c} = 1$ .

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S5 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Introduction to Logarithms, Khan Academy, embedded from YouTube.

1Which power of 2 is 1000?

2Graphs and properties of ax a^x ax

3The logarithm is the inverse of the exponential

Which power of 2 is 1000?

Graphs and properties of $a^{x}$

The logarithm is the inverse of the exponential