Exponential and Logarithmic Functions

Q: How do I sketch a log graph quickly?

Sketch $y = b^x$ first, then reflect it in $y = x$. The point $(0\,;\,1)$ becomes $(1\,;\,0)$.

A logarithm answers the question "what exponent gives this number?", which makes the log function the inverse of the exponential function. In Paper 1 you need to switch between the two forms, sketch both graphs, and read off domain, range and inequalities.

Definition of a logarithm

For $b > 0$ , $b \neq = 1$ and $x > 0$ :

$y = lo g_{b} x ⟺ x = b^{y}$

The logarithm is the exponent.

Examples.

$2^{5} = 32 ⟺ lo g_{2} 32 = 5$
$lo g_{3} 81 = 4$ because $3^{4} = 81$
$lo g_{2} \frac{1}{8} = - 3$ because $2^{- 3} = \frac{1}{8}$

$lo g x$ with no base means base 10. You cannot take the log of 0 or of a negative number.

y = log_b x as the inverse of y = b^x

Swapping $x$ and $y$ in $y = b^{x}$ gives $x = b^{y}$ , which in log form is $y = lo g_{b} x$ . So

$f (x) = b^{x} \Rightarrow f^{- 1} (x) = lo g_{b} x$

Example. If $f (x) = (\frac{1}{3})^{x}$ , then $f^{- 1} (x) = lo g_{\frac{1}{3}} x$ .

$f^{- 1} (9) = lo g_{\frac{1}{3}} 9 = - 2$ , because $(\frac{1}{3})^{- 2} = 9$ .

The graph of $y = lo g_{b} x$ is the reflection of $y = b^{x}$ in the line $y = x$ .

Graphs, domain and range

Comparing $y = b^{x}$ and $y = lo g_{b} x$ :

Exponential $y = b^{x}$ : domain all real $x$ ; range $y > 0$ ; $y$ -intercept $(0; 1)$ ; asymptote $y = 0$
Logarithmic $y = lo g_{b} x$ : domain $x > 0$ ; range all real $y$ ; $x$ -intercept $(1; 0)$ ; asymptote $x = 0$

If $b > 1$ both graphs increase. If $0 < b < 1$ both decrease.

Reading inequalities. For $g (x) = lo g_{2} x$ , $g (x) \leq 3$ means $x \leq 2^{3} = 8$ . The domain also requires $x > 0$ , so the answer is $0 < x \leq 8$ .

Key terms

Logarithm: $lo g_{b} x$ is the exponent to which $b$ must be raised to give $x$ .
Base: The number $b$ in $b^{x}$ or $lo g_{b} x$ ; it must be positive and not equal to 1.
Exponential function: A function of the form $y = b^{x}$ with the variable in the exponent.
Vertical asymptote: For $y = lo g_{b} x$ this is the $y$ -axis, $x = 0$ .

Common mistakes

Forgetting the domain restriction $x > 0$ when solving log inequalities.
Mixing up which value is the exponent when converting between forms.
Drawing the log graph crossing the $y$ -axis.
Reversing the inequality incorrectly. With base greater than 1, the direction stays the same.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

(a) Write $2^{5} = 32$ in logarithmic form. (1)

(b) Calculate $lo g_{3} 81$ without a calculator. (1)

Worked solution

Hidden - try the question first

Solution

(a) $lo g_{2} 32 = 5$
(b) $3^{4} = 81$ , so $lo g_{3} 81 = 4$
(c) $2^{- 3} = \frac{1}{8}$ , so $lo g_{2} \frac{1}{8} = - 3$

Question 2

5 marks total

routine

Given $f (x) = (\frac{1}{3})^{x}$ .

(a) Write down $f^{- 1} (x)$ in the form $y = \dots$ (2)

(b) Calculate $f^{- 1} (9)$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $lo g_{2} 32 = 5$
$f^{- 1} (x) = lo g_{\frac{1}{3}} x$
(b) $(\frac{1}{3})^{y} = 9 = (\frac{1}{3})^{- 2}$
$f^{- 1} (9) = - 2$
(c) Decreasing, because the base $\frac{1}{3}$ is between 0 and 1

Question 3

6 marks total

complex

Given $g (x) = lo g_{2} x$ .

(a) Sketch $g$ and $g^{- 1}$ on the same set of axes. Show all intercepts and one other point on each graph. (3)

(b) For which values of $x$ is $g (x) \leq 3$ ? (2)

Worked solution

Hidden - try the question first

Solution

(a) $lo g_{2} 32 = 5$ with $y$ -intercept $(0; 1)$ and point $(1; 2)$
$g$ with $x$ -intercept $(1; 0)$ and point $(2; 1)$
Correct shapes: $(\frac{1}{3})^{y} = 9 = (\frac{1}{3})^{- 2}$ approaches the $y$ -axis, $g^{- 1}$ approaches the $x$ -axis, mirror images in $y = x$
(b) $x \leq 2^{3} = 8$
and $x > 0$ , so $0 < x \leq 8$
(c) $(2; 0)$

Question 4

8 marks total

problem-solving

The graph of $f (x) = b^{x}$ passes through $(2; 16)$ .

(a) Determine the value of $b$ . (2)

(b) Write down the equation of $f^{- 1}$ . (1)

(d) For which values of $x$ is $f^{- 1} (x) > - 1$ ? (2)

(e) $g$ is the reflection of $f^{- 1}$ in the $x$ -axis. Write $g$ as a logarithm with a single base. (1)

Worked solution

Hidden - try the question first

Solution

(a) $lo g_{2} 32 = 5$
$b = 4$ (since $b > 0$ )
(b) $f^{- 1} (x) = lo g_{4} x$
(c) $x = 4^{1, 5}$
$= 8$
(d) $lo g_{4} x > - 1 \Rightarrow x > 4^{- 1}$
$x > \frac{1}{4}$
(e) $g (x) = - lo g_{4} x = lo g_{\frac{1}{4}} x$

FAQs

Do I need log laws for IEB Grade 12?

Basic log laws are useful for simplifying and for finding $n$ in finance, but the main exam focus is the definition and the link between log and exponential graphs.

Why is the domain of a log function x > 0?

Because a positive base raised to any power is always positive, so there is no exponent that gives 0 or a negative number.

How do I sketch a log graph quickly?

Sketch $y = b^{x}$ first, then reflect it in $y = x$ . The point $(0; 1)$ becomes $(1; 0)$ .

Definition of a logarithm

y = log_b x as the inverse of y = b^x

Graphs, domain and range

Key terms

Common mistakes

Practice questions

FAQs

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