IB Analysis and Approaches HL · Functions · Section 12 of 15

Exponential and logarithmic functions

3 worked examples · By Álvaro López

Key idea

The exponential $a^{x}$ grows (or decays) by multiplying by the same factor at each step, and the logarithm is its inverse function. Knowing their graphs, their domain and their asymptote lets you solve modelling problems and find parameters from two points. In the IB exam they appear in both papers, almost always in a real-life context.

The exponential function

Whatever the value of $a$ , the graph always has the same shape:

the domain is $R$ and the range is $] 0, + \infty [$ (it is always positive);
it cuts the $y$ -axis at $(0, 1)$ , because $a^{0} = 1$ , and it does not cut the $x$ -axis;
the $x$ -axis ( $y = 0$ ) is a horizontal asymptote: the graph approaches it, without touching it, as $x \to - \infty$ if $a > 1$ , or as $x \to + \infty$ if $0 < a < 1$ ;
it is increasing if $a > 1$ and decreasing if $0 < a < 1$ .

Graphs of y = 2 to the power x (increasing, orange) and of y = (1/2) to the power x (decreasing, blue); both pass through (0, 1) and approach the x-axis without touching it

The two curves are symmetric with respect to the $y$ -axis, because $(\frac{1}{2})^{x} = 2^{- x}$ : changing the sign of $x$ reflects the graph.

The number $e$

The number $e \approx 2.718$ is the "natural" base of calculus. The function $f (x) = e^{x}$ (increasing) and $f (x) = e^{- x}$ (decreasing) are the exponentials most used in models. Any $a^{x}$ can be written with base $e$ : since $a = e^{l n a}$ , we get $a^{x} = e^{x l n a}$ .

Shifted and stretched exponentials

A general exponential is $f (x) = k a^{x} + c$ , with $k \neq = 0$ . Using what you already know about transformations, $k$ stretches the graph vertically (and reflects it if it is negative), and $c$ moves it up or down.

This is the key to finding parameters: the asymptote gives $c$ , the point on the $y$ -axis gives $k$ , and another point on the graph gives $a$ .

The logarithmic function

As it is the inverse of the exponential, its graphs are symmetric with respect to the line $y = x$ : the points $(x, y)$ of one correspond to the points $(y, x)$ of the other. From this we deduce the properties of $f (x) = lo g_{a} x$ (with $a > 1$ ):

domain $] 0, + \infty [$ and range $R$ (the domain and range of the exponential, swapped);
it cuts the $x$ -axis at $(1, 0)$ , because $lo g_{a} 1 = 0$ , and it does not cut the $y$ -axis;
the $y$ -axis ( $x = 0$ ) is a vertical asymptote: $lo g_{a} x \to - \infty$ when $x \to 0^{+}$ ;
it is increasing.

The graphs of y = e to the power x (orange) and of y = ln x (blue) are symmetric with respect to the line y = x (dashed); they pass through (0, 1) and (1, 0)

If the logarithm is shifted, $ln (x - h) + d$ has vertical asymptote $x = h$ and domain $] h, + \infty [$ : the argument of the logarithm must be positive.

Worked examples

Example 1 Basic

Values and asymptote of a decreasing exponential

Let $f (x) = 4 \cdot 2^{- x} + 1$ .

a) Calculate the point at which the graph cuts the $y$ -axis (give the value of $f (0)$ ).

b) Find the equation of the horizontal asymptote (give the value of $y$ ).

c) Solve $f (x) = 3$ .

Solution

a) $f (0) = 4 \cdot 2^{0} + 1 = 5$ . The graph cuts the $y$ -axis at $(0, 5)$ .

b) As $x \to + \infty$ , $2^{- x} \to 0$ and $f (x) \to 1$ . The horizontal asymptote is $y = 1$ .

c) $4 \cdot 2^{- x} + 1 = 3$ gives $4 \cdot 2^{- x} = 2$ , that is, $2^{- x} = \frac{1}{2} = 2^{- 1}$ . Equating the exponents, $- x = - 1$ , so $x = 1$ .

Example 2 Medium

A shifted logarithmic function and its inverse

Let $f (x) = ln (x - 3) + 2$ .

a) Find the equation of the vertical asymptote (give the value of $x$ ).

b) Find the point at which the graph cuts the $x$ -axis (give the exact value of $x$ ).

c) Find $f^{- 1} (x)$ .

Solution

a) The argument of the logarithm must be positive: $x - 3 > 0$ , that is, $x > 3$ . The domain is $] 3, + \infty [$ and the vertical asymptote is $x = 3$ .

b) Solve $f (x) = 0$ :

$ln (x - 3) = - 2 ⟹ x - 3 = e^{- 2} ⟹ x = 3 + e^{- 2} \approx 3.14$

c) Write $y = ln (x - 3) + 2$ and swap the variables: $x = ln (y - 3) + 2$ . Solve for $y$ :

$x - 2 = ln (y - 3) ⟹ y - 3 = e^{x - 2} ⟹ f^{- 1} (x) = e^{x - 2} + 3$

The domain of $f^{- 1}$ is $R$ , the range of $f$ . Its horizontal asymptote $y = 3$ corresponds to the vertical asymptote $x = 3$ of $f$ .

Example 3 Hard

Parameters of an exponential from its graph

A function has the form $f (x) = k a^{x} + c$ , with $a > 0$ . Its graph passes through the points $(0, 3)$ and $(1, 5)$ and has the horizontal asymptote $y = - 1$ as $x \to - \infty$ .

a) Find the values of $k$ , $a$ and $c$ . Give the tuple $(k, a, c)$ .

b) Find the exact value of $x$ for which $f (x) = 0$ .

c) Find the exact value of $x$ for which $f (x) = 20$ .

Solution

a) The horizontal asymptote is $y = c$ , so $c = - 1$ .

The point $(0, 3)$ gives $f (0) = k + c = 3$ , that is, $k - 1 = 3$ , so $k = 4$ .

The point $(1, 5)$ gives $f (1) = k a + c = 5$ , that is, $4 a - 1 = 5$ , so $a = \frac{3}{2}$ .

Thus $f (x) = 4 (\frac{3}{2})^{x} - 1$ , which is increasing and agrees with the asymptote towards $- \infty$ .

b) $4 (\frac{3}{2})^{x} - 1 = 0$ gives $(\frac{3}{2})^{x} = \frac{1}{4}$ . Taking logarithms,

$x = \frac{ln \frac{1}{4}}{ln \frac{3}{2}} = - \frac{ln 4}{ln \frac{3}{2}} \approx - 3.42$

c) $4 (\frac{3}{2})^{x} - 1 = 20$ gives $(\frac{3}{2})^{x} = \frac{21}{4}$ , so

$x = \frac{ln \frac{21}{4}}{ln \frac{3}{2}} \approx 4.09$

In the exam: if the question asks for the exact value, leave the quotient of logarithms; if it asks for an approximation (Paper 2), round to three significant figures. Always use the condition from the asymptote before substituting the points: it simplifies the system.

Common mistakes

Summary

$f (x) = a^{x}$ ( $a > 0$ , $a \neq = 1$ ): domain $R$ , range $] 0, + \infty [$ , cuts the $y$ -axis at $(0, 1)$ , horizontal asymptote $y = 0$ ; increasing if $a > 1$ and decreasing if $0 < a < 1$ .
$f (x) = k a^{x} + c$ : horizontal asymptote $y = c$ and intercept with the $y$ -axis at $(0, k + c)$ .
$lo g_{a} x$ is the inverse of $a^{x}$ : domain $] 0, + \infty [$ , range $R$ , cuts the $x$ -axis at $(1, 0)$ , vertical asymptote $x = 0$ . $ln x$ is the inverse of $e^{x}$ .
The graphs of $f$ and $f^{- 1}$ are symmetric with respect to $y = x$ . To invert: swap $x$ and $y$ and solve for $y$ .
With $ln (x - h) + d$ the domain is $] h, + \infty [$ and the vertical asymptote is $x = h$ .

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Exponential and logarithmic functions

The exponential function

The number e

Shifted and stretched exponentials

The logarithmic function

Worked examples

Common mistakes

Summary

Want to work on it together?

The number $e$