IB Analysis and Approaches HL · Exponential equations and logarithms · Section 1 of 3

Exponential equations

3 worked examples · By Álvaro López

Key idea

An exponential equation is one in which the unknown appears in an exponent. In this section you will see the three methods that are solved without logarithms: equating the bases, taking out a common factor and the substitution $t = a^{x}$ . These are the methods asked for in Paper 1 (no calculator) and the basis for the sections that follow.

What an exponential equation is

Neither $x^{2} = 9$ nor $x^{3} = 8$ is one: there the unknown is in the base. Here, by contrast, the base is fixed and what you are looking for is the exponent.

Before solving, it helps to recall the laws of powers, because almost all the work consists of applying them correctly. For $a > 0$ :

$a^{m} \cdot a^{n} = a^{m + n} \frac{a ^{m}}{a ^{n}} = a^{m - n} (a^{m})^{n} = a^{mn} a^{- n} = \frac{1}{a ^{n}} a^{1/ n} = n a$

Method 1: equating the bases

A power with base $a > 0$ , where $a \neq = 1$ , takes each value only once: if the exponent changes, the result changes. This lets you "remove" the bases when they are equal.

The reason is that $a^{x}$ never repeats a value. For example, $2^{3} = 8$ and $2^{4} = 16$ ; no other exponent gives $8$ . The condition $a \neq = 1$ is necessary: $1^{u} = 1^{v}$ holds always, whatever $u$ or $v$ are.

The problem is that the equation almost never comes with equal bases. You have to write each side as a power of the same base, usually a small number (2, 3, 5, 10…). The usual clues:

A number such as $32$ , $\frac{1}{8}$ or $2$ is a power of $2$ : $2^{5}$ , $2^{- 3}$ , $2^{1/2}$ .
A fraction $(\frac{1}{4})^{x}$ is brought to base $2$ with a negative exponent: $(\frac{1}{4})^{x} = 2^{- 2 x}$ .
A power of a power is multiplied: $(9^{x})^{2} = 3^{4 x}$ .

Method 2: taking out a common factor

When there are several terms with exponents of the type $x + 1$ , $x + 2$ , $x - 1$ … and the same base, you cannot equate the bases yet (there is a sum). What you do is separate the factor $a^{x}$ using the laws of the product:

$2^{x + 3} = 2^{x} \cdot 2^{3} = 8 \cdot 2^{x} 5^{x - 1} = \frac{5 ^{x}}{5}$

This way every term becomes a number multiplied by $a^{x}$ , and you take out the common factor. For example, $2^{x + 1} + 2^{x} = 2 \cdot 2^{x} + 2^{x} = 3 \cdot 2^{x}$ .

You can write $t = a^{x}$ from the start to work with fewer symbols: the equation becomes $c t = d$ and at the end you undo the substitution.

Method 3: substitution $t = a^{x}$

If the equation mixes $a^{2 x}$ (or $(a^{x})^{2}$ ) with $a^{x}$ and numbers, there is no way to take out a common factor, but you can see it as a quadratic equation. Since $a^{2 x} = (a^{x})^{2}$ , putting $t = a^{x}$ gives a polynomial in $t$ .

This is the part that scores the most marks and is most often forgotten: the equation in $t$ may have two solutions and only one of them gives values of $x$ .

If $t$ is in a denominator (for example, $2^{- x} = \frac{1}{2 ^{x}} = \frac{1}{t}$ ), multiply by $t$ to remove it; there is no problem, because $t > 0$ is never zero.

When the bases cannot be equated

Not every equation can be solved with the methods above. In $2^{x} = 5$ , the number $5$ is neither an integer nor a fractional power of $2$ , and there is no way to equate the bases. A solution exists, a number between $2$ and $3$ (because $2^{2} = 4 < 5 < 8 = 2^{3}$ ), but to write it exactly you need a new tool: logarithms, which are studied in the next section.

Worked examples

Example 1 Basic

Equating bases with a fraction

Solve $(\frac{1}{4})^{x - 2} = 32$ .

Solution

Common base. $\frac{1}{4} = 2^{- 2}$ and $32 = 2^{5}$ . Both quantities are powers of $2$ .
Rewrite each side as a power of $2$ :

$(2^{- 2})^{x - 2} = 2^{5} ⟹ 2^{- 2 (x - 2)} = 2^{5}$

Equate the exponents:

$- 2 (x - 2) = 5 ⟺ - 2 x + 4 = 5 ⟺ x = - \frac{1}{2}$

Check. With $x = - \frac{1}{2}$ : $x - 2 = - \frac{5}{2}$ and $(\frac{1}{4})^{- 5/2} = 4^{5/2} = (4)^{5} = 2^{5} = 32$ .

Example 2 Medium

Common factor and substitution

Solve each equation.

a) $3^{x + 2} - 3^{x} = 216$

b) $4^{x} - 3 \cdot 2^{x} - 4 = 0$

Solution

a) 1. Take out the factor $3^{x}$ . $3^{x + 2} = 3^{x} \cdot 3^{2} = 9 \cdot 3^{x}$ , so

$9 \cdot 3^{x} - 3^{x} = 216 ⟺ 8 \cdot 3^{x} = 216 ⟺ 3^{x} = 27$

Equate bases. $27 = 3^{3}$ , so $x = 3$ .
Check. $3^{5} - 3^{3} = 243 - 27 = 216$ .

b) 1. Write everything in base $2$ . $4^{x} = (2^{2})^{x} = (2^{x})^{2}$ . The equation is $(2^{x})^{2} - 3 \cdot 2^{x} - 4 = 0$ .

Substitution. With $t = 2^{x}$ :

$t^{2} - 3 t - 4 = 0 ⟺ (t - 4) (t + 1) = 0 ⟺ t = 4 or t = - 1$

Reject. Since $2^{x} > 0$ for every $x$ , the solution $t = - 1$ is not valid: $2^{x} = - 1$ has no solution.
Undo the substitution. $2^{x} = 4 = 2^{2}$ , so $x = 2$ .
Check. $4^{2} - 3 \cdot 2^{2} - 4 = 16 - 12 - 4 = 0$ .

The quadratic equation had two solutions, but the exponential equation has only one.

Example 3 Hard

A parameter and the number of solutions

Consider the equation $2^{x} + 2^{3 - x} = k$ , where $k$ is a positive constant.

a) For $k = 9$ , find all the values of $x$ that satisfy the equation.

b) Find the value of $k$ for which the equation has exactly one solution.

c) For that value of $k$ , find the solution.

Solution

Substitution (valid for any $k$ ). Since $2^{3 - x} = \frac{2 ^{3}}{2 ^{x}} = \frac{8}{2 ^{x}}$ , with $t = 2^{x} > 0$ the equation is

$t + \frac{8}{t} = k$

We multiply by $t$ , which is not zero, and rearrange:

$t^{2} - k t + 8 = 0$

a) With $k = 9$ : $t^{2} - 9 t + 8 = 0 ⟺ (t - 1) (t - 8) = 0$ , so $t = 1$ or $t = 8$ . Both are positive and valid.

$2^{x} = 1 = 2^{0} \Rightarrow x = 0$ .
$2^{x} = 8 = 2^{3} \Rightarrow x = 3$ .

Check: $2^{0} + 2^{3} = 1 + 8 = 9$ and $2^{3} + 2^{0} = 8 + 1 = 9$ . Solutions: $x = 0$ and $x = 3$ .

b) Each $t > 0$ gives exactly one value of $x$ . The equation in $t$ must therefore have exactly one positive root, that is, a double root: the discriminant is zero.

$Δ = k^{2} - 4 \cdot 1 \cdot 8 = k^{2} - 32 = 0 ⟺ k = \pm 42$

Since $k > 0$ , $k = 42$ . (The double root is $t = \frac{k}{2} = 22 > 0$ , so it is valid.)

c) With $k = 42$ : $t = 22 = 2 \cdot 2^{1/2} = 2^{3/2}$ . Then $2^{x} = 2^{3/2}$ and $x = \frac{3}{2}$ .

Check: $2^{3/2} + 2^{3 - 3/2} = 2^{3/2} + 2^{3/2} = 2 \cdot 22 = 42$ .

In the exam: marks are given for the set-up $t^{2} - k t + 8 = 0$ , for justifying that $t > 0$ and for explaining why a single solution requires a zero discriminant. Do not forget to undo the substitution in c): $t = 22$ is not the value of $x$ .

Common mistakes

Summary

Exponential equation: the unknown is in the exponent. For $a > 0$ and $a \neq = 1$ , $a^{u} = a^{v} ⟺ u = v$ .
Equating bases: write both sides as a power of the same base (for example $\frac{1}{4} = 2^{- 2}$ ) and equate the exponents.
Common factor: with exponents $x + k$ , use $a^{x + k} = a^{k} a^{x}$ and leave $c \cdot a^{x} = d$ .
Substitution: $t = a^{x}$ turns $a^{2 x}$ and $a^{x}$ into a quadratic in $t$ ; only the solutions $t > 0$ are valid.
If the bases cannot be equated (such as $2^{x} = 5$ ), use logarithms: next section.

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Exponential equations

What an exponential equation is

Method 1: equating the bases

Method 2: taking out a common factor

Method 3: substitution t=ax

When the bases cannot be equated

Worked examples

Common mistakes

Summary

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Method 3: substitution $t = a^{x}$