International Baccalaureate · MYP 5 Extended

Exponents and Exponential Functions

⌚ ~186 min ▢ 21 blocks ✓ 97 exercises ➣ Prerequisites : Functions, Exponents and Standard Form, Radicals and Surds, Number Sequences and Finance

I Exponents

I.1 Rational Exponents

Discover

We know about positive exponents, like $5^3 = 5 \times 5 \times 5$, and also about negative exponents, like $5^{-3} = \dfrac{1}{5 \times 5 \times 5}$.
But what about fractional exponents?
Using the exponent laws, look at what happens with $\textcolor{colordef}{5^{\frac{1}{2}}}$:$$\begin{aligned}\textcolor{colordef}{5^{\frac{1}{2}}} \times \textcolor{colordef}{5^{\frac{1}{2}}} &= 5^{\frac{1}{2}+\frac{1}{2}} \\ &= 5^1 \\ &= \textcolor{olive}{5}\end{aligned}$$ And by the definition of the square root:

$$\textcolor{colorprop}{\sqrt{5}} \times \textcolor{colorprop}{\sqrt{5}} = \textcolor{olive}{5}$$ By comparing these two results, we see that:$$\textcolor{colordef}{5^{\frac{1}{2}}} \times \textcolor{colordef}{5^{\frac{1}{2}}} = \textcolor{colorprop}{\sqrt{5}} \times \textcolor{colorprop}{\sqrt{5}}$$ So, we can deduce that:$$\textcolor{colordef}{5^{\frac{1}{2}}} = \textcolor{colorprop}{\sqrt{5}}$$ Rational exponents are used only with positive bases, so that roots like $\sqrt[n]{a}$ are real numbers. This shows us that we can use fractional exponents to represent roots, extending our understanding of exponents to include rational exponents.

Definition — Rational Exponent

For a positive number $a$ and positive integers $m$ and $n$, $$\begin{aligned} a^{\frac{1}{2}} &= \sqrt{a}, \\ a^{\frac{1}{n}} &= \sqrt[n]{a}, \\ a^{\frac{m}{n}} &= \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m. \end{aligned}$$

For a positive base $a$, the exponent laws known for integer exponents remain true for rational exponents; they are used here without proof, as the opening problem of this section did with $5^{\frac{1}{2}} \times 5^{\frac{1}{2}} = 5^{\frac{1}{2} + \frac{1}{2}}$. In particular, $a^{-p} = \dfrac{1}{a^{p}}$ for every rational number $p$.

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Example

Without a calculator, evaluate:

$8^{\frac{2}{3}}$;
$16^{-\frac{3}{4}}$.

Answer

The definition gives two ways to write $a^{\frac{m}{n}}$. The form $\left(\sqrt[n]{a}\right)^m$ takes the root first, so the numbers stay small.

$$\begin{aligned}[t] 8^{\frac{2}{3}} &= \left(\sqrt[3]{8}\right)^2 && \text{(definition of a rational exponent)} \\ &= 2^2 && \text{(since } 2^3 = 8\text{)} \\ &= 4. \end{aligned}$$
$$\begin{aligned}[t] 16^{-\frac{3}{4}} &= \dfrac{1}{16^{\frac{3}{4}}} && \text{(negative exponent: } a^{-p} = \tfrac{1}{a^{p}}\text{)} \\ &= \dfrac{1}{\left(\sqrt[4]{16}\right)^3} && \text{(definition of a rational exponent)} \\ &= \dfrac{1}{2^3} && \text{(since } 2^4 = 16\text{)} \\ &= \dfrac{1}{8}. \end{aligned}$$

Taking the power first also works, but with larger numbers: $8^{\frac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$.

Skills to practice

Expressing Roots Using Exponents
Evaluating Rational Powers without a Calculator
Calculating Powers and Rounding
Simplifying Expressions with Rational Exponents

II Exponential Functions

II.1 The Exponential Function and Its Graph

Expressions like $a^n$ make sense when the exponent $n$ is an integer (roots give fractional exponents). The exponential function, written $f(x) = a^x$, extends that concept to any real exponent $x$.
This chapter presents the key features and graphs of these functions, and how they model real-world phenomena involving rapid growth or decay, such as population dynamics and compound interest.

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Definition — Exponential Function

The exponential function has the form $f(x)=a^x$, where the base $a$ is a positive constant and $a \neq 1$.

Proposition — Key Features of the Graph of $y\equal a^x$

All exponential functions of the form $f(x) = a^x$ share several key graphical features:

Domain: The domain is all real numbers, $(-\infty, \infty)$.
Range: The range is all positive real numbers, $(0, \infty)$.
Horizontal Asymptote: The graph has a horizontal asymptote at the x-axis ($y=0$). The function approaches this line but never touches it.
y-intercept: The graph always passes through the point $(0, 1)$, because $a^0 = 1$ for any valid base $a$.
General Shape: The shape of the graph is determined by the value of the base, $a$:
- If $a > 1$, the function shows exponential growth and is increasing.
- If $0 < a < 1$, the function shows exponential decay and is decreasing.

$\quad$

Proposition — Algebraic Properties of $x \mapsto a^x$

Let $a \gt 0$. The rules of integer powers remain true for every real exponent; these properties are accepted without proof. For all real numbers $x$ and $y$ and every integer $n$:

$\textcolor{colorprop}{a^{x+y} = a^x \times a^y}$;
$\textcolor{colorprop}{a^{x-y} = \dfrac{a^x}{a^y}}$, and in particular $\textcolor{colorprop}{a^{-x} = \dfrac{1}{a^x}}$;
$\textcolor{colorprop}{a^{nx} = \left(a^x\right)^n}$.

They are used to transform numerical or literal expressions — the same gestures as with integer exponents.

Example

Write each expression as a single power, or as a number.

$2^{1.5} \times 2^{0.5}$
$\dfrac{5^{x+1}}{5^{x}}$
$\left(1.05^{x}\right)^{12}$
$\dfrac{1}{2^{x}}$

Answer

$\begin{aligned}[t]2^{1.5} \times 2^{0.5} &= 2^{1.5 + 0.5} &&\text{(product: add the exponents)}\\ &= 2^{2}\\ &= 4\end{aligned}$
$\begin{aligned}[t]\dfrac{5^{x+1}}{5^{x}} &= 5^{(x+1) - x} &&\text{(quotient: subtract the exponents)}\\ &= 5^{1}\\ &= 5\end{aligned}$
$\begin{aligned}[t]\left(1.05^{x}\right)^{12} &= 1.05^{12x} &&\text{(} a^{nx} = (a^x)^n \text{ with } n = 12\text{)}\end{aligned}$
$\begin{aligned}[t]\dfrac{1}{2^{x}} &= 2^{-x} &&\text{(} a^{-x} = \tfrac{1}{a^x}\text{)}\end{aligned}$

Proposition — Variations of $x \mapsto k \times a^x$

Let $a \gt 0$ with $a \neq 1$, and let $k \neq 0$. The function $f$ defined for every real $x$ by $f(x) = k \times a^x$ satisfies $f(0) = k$ and has the sign of $k$.

If $\textcolor{colorprop}{k \gt 0}$, $f$ has the same variations as $x \mapsto a^x$: increasing if $a \gt 1$, decreasing if $0 \lt a \lt 1$.
If $\textcolor{colorprop}{k \lt 0}$, $f$ has the opposite variations: decreasing if $a \gt 1$, increasing if $0 \lt a \lt 1$.

(For $a = 1$ or $k = 0$, $f$ is constant.) Multiplying by a positive number keeps the order of two numbers; multiplying by a negative number reverses it. At the whole numbers $n = 0, 1, 2, \ldots$, the values $k \times a^n$ are multiplied by $a$ at each step: they form a geometric sequence, with first term $k$ and common ratio $a$, and the function and the sequence vary in the same way.

Example

Give the variations of each function on the set of real numbers.

$f(x) = 3 \times 1.2^{x}$
$g(x) = -2 \times 1.5^{x}$
$h(x) = 5 \times 0.8^{x}$
$m(x) = -0.5 \times 0.9^{x}$

Answer

$k = 3 \gt 0$ and $a = 1.2 \gt 1$: $f$ is increasing.
$k = -2 \lt 0$ and $a = 1.5 \gt 1$: $g$ is decreasing.
$k = 5 \gt 0$ and $0 \lt a = 0.8 \lt 1$: $h$ is decreasing.
$k = -0.5 \lt 0$ and $0 \lt a = 0.9 \lt 1$: $m$ is increasing — two reversals cancel out.

Proposition — Inserting a Point on the Curve

Let $a \gt 0$ with $a \neq 1$, let $k \gt 0$, and let $f(x) = k \times a^x$. For any two real numbers $x_1$ and $x_2$: $$\textcolor{colorprop}{f\left(\frac{x_1 + x_2}{2}\right) = \sqrt{f(x_1) \times f(x_2)}}.$$ On the curve of $f$, the point whose $x$-coordinate is the arithmetic mean of two $x$-coordinates has as its $y$-coordinate the geometric mean of the two $y$-coordinates (the positive number whose square is their product).

Proof

Let $m = \dfrac{x_1 + x_2}{2}$, so that $2m = x_1 + x_2$. Using the algebraic properties of $x \mapsto a^x$: $$\begin{aligned} f(m)^2 &= \left(k \times a^{m}\right)^2 \\ &= k^2 \times \left(a^{m}\right)^2 && \text{(square of a product)}\\ &= k^2 \times a^{2m} && \text{($a^{nx} = \left(a^x\right)^n$ with $n = 2$)}\\ &= k^2 \times a^{x_1 + x_2} && \text{($2m = x_1 + x_2$)}\\ &= \left(k \times a^{x_1}\right) \times \left(k \times a^{x_2}\right) && \text{($a^{x+y} = a^x \times a^y$)}\\ &= f(x_1) \times f(x_2). \end{aligned}$$ Since $k \gt 0$ and $a^m \gt 0$, $f(m) \gt 0$: it is the positive number whose square is $f(x_1) \times f(x_2)$, that is $f(m) = \sqrt{f(x_1) \times f(x_2)}$.

Method — Building the Curve Point by Point

To place points of the curve of $f(x) = k \times a^x$ ($k \gt 0$) without computing any power with a non-integer exponent:

Start from two points of the curve already known, for example $A(0, k)$ and $B(1, k \times a)$.
Between two neighbouring points $(x_1, y_1)$ and $(x_2, y_2)$, insert the point $\left(\dfrac{x_1 + x_2}{2}, \sqrt{y_1 \times y_2}\right)$: it is on the curve.
Repeat step 2 between all the neighbouring points: each round doubles the number of intervals, and the points fill the arc of the curve between $A$ and $B$ more and more densely.

Each round applies the same step to every pair of neighbouring points: it is an algorithm, which can be repeated as many times as wanted.

Example

The points $A(0, 1)$ and $B(1, 4)$ are on the curve of $f(x) = 4^x$. Carry out two rounds of insertion between $A$ and $B$, give the coordinates of the points obtained, and check one of them with the power key of a calculator.

Answer

First round. Between $A(0, 1)$ and $B(1, 4)$: $x$-coordinate $\dfrac{0 + 1}{2} = 0.5$, $y$-coordinate $\sqrt{1 \times 4} = 2$; the point $(0.5, 2)$. Second round. Between $(0, 1)$ and $(0.5, 2)$: $(0.25, \sqrt{2}) \approx (0.25, 1.414)$. Between $(0.5, 2)$ and $(1, 4)$: $(0.75, \sqrt{8}) \approx (0.75, 2.828)$. Check: the calculator gives $4^{0.75} \approx 2.828$. The five points are on the curve:

Skills to practice

Recognising the Variations of $k \times a^x$
Reading and Evaluating Exponential Functions
Using the Algebraic Properties of $x \mapsto a^x$
Sketching and Studying Exponential Curves

II.2 Transformations of Exponential Graphs

Proposition — Transformations of Exponential Functions

The graph of the general exponential function $f(x) = k \cdot a^{m(x-h)} + v$, with $a > 0$, $a \neq 1$, $k \neq 0$ and $m \neq 0$, can be obtained by applying transformations to the basic graph of $y=a^x$.

Vertical Translation ($v$): The graph is shifted vertically by $|v|$ units: up if $v > 0$, down if $v < 0$. The horizontal asymptote becomes $y=v$.
Horizontal Translation ($h$): The graph is shifted horizontally by $|h|$ units: to the right if $h > 0$, to the left if $h < 0$.
Vertical Stretch/Reflection ($k$): The graph is stretched vertically by a factor of $|k|$. If $k<0$, the graph is reflected in the horizontal asymptote.
Horizontal Stretch/Reflection ($m$): The graph is stretched horizontally by a factor of $1/|m|$ about the line $x=h$. If $m<0$, it is reflected in the vertical line $x=h$ (in the $y$-axis only when $h=0$).

Example

The curve of $y = 2^{x-1} - 3$ is obtained from the curve of $y = 2^{x}$. Describe the transformation, give the equation of the horizontal asymptote and the $y$-intercept, then sketch the curve.

Answer

Transformation. The point $(x, y)$ of the curve of $y = 2^{x}$ moves to the point $(x + 1, y - 3)$: for the $x$-coordinate $x + 1$, the exponent $(x + 1) - 1 = x$ gives the same power $2^{x}$, then $3$ is subtracted. The curve of $y = 2^{x}$ is translated $1$ unit to the right and $3$ units down. For example, $(0, 1)$ moves to $(1, -2)$ and $(2, 4)$ moves to $(3, 1)$. Asymptote. The asymptote $y = 0$ of the curve of $y = 2^{x}$ also moves $3$ units down: the horizontal asymptote is $\boldsymbol{y = -3}$. $y$-intercept. For $x = 0$: $y = 2^{0-1} - 3 = \dfrac{1}{2} - 3 = \boldsymbol{-\dfrac{5}{2}}$; the curve crosses the $y$-axis at $\left(0, -\dfrac{5}{2}\right)$.

Skills to practice

Describing and Studying Transformed Exponential Functions

III Exponential Growth and Decay

III.1 Linear or Exponential?

Many real-world patterns of growth and decay can be modelled by either linear or exponential functions. The key to distinguishing between them is understanding how the quantity changes over regular intervals.

Linear relationships involve additive change. The same amount is added or subtracted in each time step. Think of saving $\dollar 50$ every month.
Exponential relationships involve multiplicative change. The amount is multiplied by the same factor in each time step. Think of a population that doubles every year.

This fundamental difference can be identified mathematically by checking for a constant difference or a constant ratio.

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Proposition — Identifying Linear vs. Exponential Relationships

For a set of data where the independent variable ($x$) increases by constant intervals:

If the difference between consecutive dependent variable ($y$) values is constant, a linear model fits the data.
If the $y$-values are non-zero and the ratio of consecutive $y$-values is constant and different from $1$, an exponential model fits the data.

A table shows only some values: it shows that a model is compatible with them. The proof below gives the other direction: a linear relationship always gives a constant difference, and an exponential relationship a constant ratio.

Proof

Linear Relationship:
Let the linear relationship be $y = ax + b$.
Consider two points, $(x_1, y_1)$ and $(x_2, y_2)$, such that the difference in their $x$-values is a constant, $k$.
So, $x_2 - x_1 = k$.
The corresponding $y$-values are $y_1 = ax_1 + b$ and $y_2 = ax_2 + b$.
The difference between these $y$-values is:$$ \begin{aligned}[t]y_2 - y_1 &= (ax_2 + b) - (ax_1 + b) \\ &= ax_2 - ax_1 \\ &= a(x_2 - x_1) \\ &= ak\end{aligned} $$ Since both $a$ (the gradient) and $k$ (the change in $x$) are constants, the difference $y_2 - y_1$ is also a constant.
Exponential Relationship:
Let the exponential relationship be $y = c \cdot a^x$, with $c \neq 0$, $a > 0$ and $a \neq 1$.
Consider two points, $(x_1, y_1)$ and $(x_2, y_2)$, such that the difference in their $x$-values is a constant, $k$.
So, $x_2 - x_1 = k$.
The corresponding $y$-values are $y_1 = c \cdot a^{x_1}$ and $y_2 = c \cdot a^{x_2}$.
The ratio of these $y$-values is:$$ \begin{aligned}[t]\frac{y_2}{y_1} &= \frac{c \cdot a^{x_2}}{c \cdot a^{x_1}} \\ &= a^{x_2 - x_1} \\ &= a^k\end{aligned} $$ Since both $a$ (the base) and $k$ (the change in $x$) are constants, the ratio $\frac{y_2}{y_1}$ is also a constant. This constant is often called the common ratio.

Method — Analysing Data in a Table

To determine if a relationship is linear or exponential from a table of values:

Ensure the $x$-values increase by a constant step.
Check for a common difference: Calculate the difference between consecutive $y$-values ($y_2 - y_1$, $y_3 - y_2$, etc.). If this value is constant, the relationship is linear.
Check for a common ratio: If the difference is not constant, calculate the ratio of consecutive $y$-values ($\frac{y_2}{y_1}$, $\frac{y_3}{y_2}$, etc.). If this value is constant, the relationship is exponential.

Example

Consider the relationship represented by this table:

$x$	0	1	2	3
$y$	5	8	11	14

Is the relationship linear or exponential?

Answer

The $x$-values increase by a constant step of 1.
We check for a common difference between consecutive $y$-values:

$y_2 - y_1 = 8 - 5 = \boldsymbol{3}$
$y_3 - y_2 = 11 - 8 = \boldsymbol{3}$
$y_4 - y_3 = 14 - 11 = \boldsymbol{3}$

Since the difference is constant, the relationship is linear.$$\begin{aligned}x: &\ 0 & \textcolor{colordef}{\xrightarrow{\;+1\;}} & 1 & \textcolor{colordef}{\xrightarrow{\;+1\;}} & 2 & \textcolor{colordef}{\xrightarrow{\;+1\;}} & 3\\ y: &\ 5 & \textcolor{colorprop}{\xrightarrow{\;+3\;}} & 8 & \textcolor{colorprop}{\xrightarrow{\;+3\;}} & 11 & \textcolor{colorprop}{\xrightarrow{\;+3\;}} & 14\\ \end{aligned}$$

Example

Consider the relationship represented by this table:

$x$	0	1	2	3
$y$	2	6	18	54

Is the relationship linear or exponential?

Answer

The $x$-values increase by a constant step of 1.
First, we check for a common difference: $6-2=4$, but $18-6=12$. The difference is not constant, so the relationship is not linear.
Next, we check for a common ratio between consecutive $y$-values:

$\frac{y_2}{y_1} = \frac{6}{2} = \boldsymbol{3}$
$\frac{y_3}{y_2} = \frac{18}{6} = \boldsymbol{3}$
$\frac{y_4}{y_3} = \frac{54}{18} = \boldsymbol{3}$

Since the ratio is constant, the relationship is exponential.$$\begin{aligned}x: &\ 0\ & \textcolor{colordef}{\xrightarrow{\;+1\;}} &\ 1\ & \textcolor{colordef}{\xrightarrow{\;+1\;}} &\ 2\ & \textcolor{colordef}{\xrightarrow{\;+1\;}} &\ 3 \\ y: &\ 2\ & \textcolor{colorprop}{\xrightarrow{\;\times 3\;}} &\ 6\ & \textcolor{colorprop}{\xrightarrow{\;\times 3\;}} &\ 18\ & \textcolor{colorprop}{\xrightarrow{\;\times 3\;}} &\ 54 \\ \end{aligned}$$

Skills to practice

Recognising Linear or Exponential Relationships from Tables
Recognising Linear or Exponential Relationships in Context

III.2 Exponential Models

Exponential functions are used to model quantities that change by a constant multiplicative factor over equal intervals of time. This core principle distinguishes them from linear functions, which change by a constant difference (addition or subtraction).
There are two main types of exponential models:

Exponential Growth: The quantity increases by a constant factor greater than 1. This is seen in phenomena like population growth and compound interest.
Exponential Decay: The quantity decreases by a constant factor between 0 and 1. This is seen in phenomena like radioactive decay and asset depreciation.

These models can be discrete, occurring in distinct steps (e.g., interest compounded annually), or continuous, occurring smoothly over time (e.g., bacterial growth).

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Definition — General Model for Exponential Growth and Decay

An exponential relationship is described by the function:$$ A(t) = A_0 \times R^t $$ where:

$A(t)$ is the amount at time $t$.
$A_0$ is the initial amount (the amount at $t=0$).
$R$ is the constant growth or decay factor per unit of time, with $R > 0$: $R = 1 + r$ for a growth at the rate $r$, $R = 1 - r$ for a decay at the rate $r$.
$t$ is the time elapsed.

The function $A(t) = A_0 \times R^t$ is a function $k \times a^x$, with $k = A_0$ and $a = R$.

Example

The population of foxes, $P$, in a specified area, $t$ years after observation began, is modelled by the equation: $P(t)=300(1.25)^t$.

How many foxes are there initially?
What is the annual percentage growth rate?
How many foxes are there after 5 years?

Answer

The initial population corresponds to $t=0$. $$P(0) = 300(1.25)^0 = 300 \times 1 = \boldsymbol{300} \text{ foxes}$$
The growth factor is $R=1.25$. Since $R = 1+r$, we have $1.25 = 1+r$, which gives $r=0.25$.
The annual growth rate is $\boldsymbol{25\pourcent}$.
Substitute $t=5$ into the equation. Since the population must be a whole number, we round to the nearest fox. $$P(5) = 300(1.25)^5 \approx 915.52... \approx \boldsymbol{916} \text{ foxes}$$

Example

An amount of $\dollar 5\,000$ is invested at $6\pourcent$ p.a. compounded annually.

Find a model for the amount, $A$, after $t$ years.
Find the amount after 4 years.

Answer

The initial amount is $A_0 = 5\,000$. The annual interest rate is $r = 0.06$.
The growth factor is $R = 1+r = 1+0.06 = 1.06$.
The model is $\boldsymbol{A(t) = 5\,000(1.06)^t}$.
After 4 years, the amount is: $$A(4) = 5\,000(1.06)^4 \approx 6\,312.38...$$ The amount is $\boldsymbol{\dollar 6\,312.38}$.

Skills to practice

Modelling Real Situations with Exponential Functions

IV Exponential Expressions and Equations

IV.1 Exponential Expressions

Definition — Exponential Expression

An exponential expression is a mathematical expression where a variable appears in the exponent. The base is usually a positive constant different from $1$.

Example

$2^x$ and $5^{x+1}$ are exponential expressions. This is different from a polynomial expression like $x^2$, where the variable is in the base, not in the exponent.

Proposition — Exponent Law 4

For non-zero numbers $a$ and $b$, and any integer $n$, $$\left(\textcolor{colordef}{a}\textcolor{colorprop}{b}\right)^{\textcolor{olive}{n}} = \textcolor{colordef}{a}^{\textcolor{olive}{n}}\, \textcolor{colorprop}{b}^{\textcolor{olive}{n}}$$

Proposition — Exponent Law 5

For non-zero numbers $a$ and $b$, and any integer $n$, $$\left(\dfrac{\textcolor{colordef}{a}}{\textcolor{colorprop}{b}}\right)^{\textcolor{olive}{n}} = \dfrac{\textcolor{colordef}{a}^{\textcolor{olive}{n}}}{\textcolor{colorprop}{b}^{\textcolor{olive}{n}}}$$

Method — Manipulating Exponential Expressions

Using the laws of exponents (index laws), we can simplify, expand, and factorise expressions where the variable is in the exponent. These skills are fundamental for solving exponential equations.

Example

Simplify $\displaystyle\frac{2^{x+1}+2^x}{2^x}$.

Answer

$$\begin{aligned}[t] \frac{2^{x+1}+2^x}{2^x} &= \frac{2^x \cdot 2^1 + 2^x}{2^x} && \text{(using the law }a^{x+1}=a^x\cdot a^1\text{)} \\ &= \frac{2^x(2+1)}{2^x} && \text{(factorise by the common factor } 2^x\text{)} \\ &= 3 && \text{(simplify by } 2^x\text{)} \end{aligned}$$

Example

Factorise $4^{x} - 5 \times 2^{x} + 4$.

Answer

The first term is a power of $2^{x}$: $4^{x} = \left(2^{2}\right)^{x} = 2^{2x} = \left(2^{x}\right)^{2}$. The expression is therefore a quadratic expression in $X = 2^{x}$: $$\begin{aligned}[t] 4^{x} - 5 \times 2^{x} + 4 &= \left(2^{x}\right)^{2} - 5 \times 2^{x} + 4 &&\text{(} a^{nx} = \left(a^x\right)^n \text{ with } n = 2\text{)}\\ &= X^{2} - 5X + 4 &&\text{(} X = 2^{x}\text{)}\\ &= (X - 1)(X - 4) &&\text{(} -1 \text{ and } -4 \text{: sum } -5\text{, product } 4\text{)}\\ &= \left(2^{x} - 1\right)\left(2^{x} - 4\right). \end{aligned}$$ Check by expanding: $\left(2^{x} - 1\right)\left(2^{x} - 4\right) = \left(2^{x}\right)^{2} - 4 \times 2^{x} - 2^{x} + 4 = 4^{x} - 5 \times 2^{x} + 4$.

Skills to practice

Simplifying Using Exponent Laws
Simplifying Exponential Expressions
Expanding Exponential Expressions
Factorising Exponential Expressions

IV.2 Exponential Equations

Definition — Exponential Equation

An exponential equation is an equation in which the unknown occurs in the exponent.

Example

$2^x = 8$ and $30 \times 3^x = 7$ are both exponential equations.

There are several methods we can use to solve exponential equations. These include solving by graphing, using technology (such as a calculator or graphing software), and using logarithms. We focus on equations that can be solved algebraically by rewriting both sides with the same base and then equating the exponents.

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Method — Solving by Equating Exponents

For $a>0$, $a\neq 1$, we have $a^x = a^y$ if and only if $x = y$. So, when both sides of an equation can be written with the same base $a$, we can equate the exponents.

Example

Solve for $x$: $2^x = 16$

Answer

$$\begin{aligned}[t] && 2^x &= 16 \\ \Leftrightarrow && 2^x &= 2^4 && \text{(Write $16$ as a power of $2$)} \\ \Leftrightarrow && x &= 4 && \text{(Equate the exponents)} \end{aligned}$$

Example

Solve the equation $9^{x} = 27^{x-1}$.

Answer

Both bases are powers of $3$: $9 = 3^{2}$ and $27 = 3^{3}$. $$\begin{aligned}[t] && 9^{x} &= 27^{x-1}\\ \Leftrightarrow && \left(3^{2}\right)^{x} &= \left(3^{3}\right)^{x-1} &&\text{(write both sides as powers of } 3\text{)}\\ \Leftrightarrow && 3^{2x} &= 3^{3x-3} &&\text{(power of a power)}\\ \Leftrightarrow && 2x &= 3x - 3 &&\text{(equate the exponents)}\\ \Leftrightarrow && x &= 3 \end{aligned}$$ Equating the exponents is valid because $x \mapsto 3^{x}$ is strictly increasing (base $3 \gt 1$): two different exponents give two different powers. Check: $9^{3} = 729$ and $27^{3-1} = 27^{2} = 729$.

Skills to practice

Solving by Equating Exponents
Solving Harder Exponential Equations

V Going further

Skills to practice

Diagnosing and Correcting
Solving Problems
Investigating Patterns in Exponential Functions
Quiz

\(x\)	0	1	2	3
\(y\)	5	8	11	14

\(x\)	0	1	2	3
\(y\)	2	6	18	54