Worked examples · Sheet 3

Exponential equations with substitution

Theory used: Substitution method Equality of powers with the same base

Exercise 1

Solve the equation

4^{x} - 5 \cdot 2^{x} + 4 = 0.

Setting $t = 2^{x}$ , with $t > 0$ , since $4^{x} = (2^{x})^{2} = t^{2}$ :

t^{2} - 5 t + 4 = 0 ⟺ t = 1 or t = 4

2^{x} = 1 = 2^{0} ⟺ x = 0, 2^{x} = 4 = 2^{2} ⟺ x = 2

Solutions: $x = 0$ and $x = 2$ .

The equation contains $4^{x} = (2^{x})^{2}$ and $2^{x}$ , so the substitution $t = 2^{x}$ transforms it into a second-degree equation in $t$ , with the condition $t > 0$ because a power with a positive base is positive. The solutions $t = 1$ and $t = 4$ are both positive, therefore both are acceptable. Each one gives an equation with the same base: from $2^{x} = 2^{0}$ it follows that $x = 0$ and from $2^{x} = 2^{2}$ it follows that $x = 2$ , by the equality of the exponents.

Exercise 2

Solve the equation

2^{2 x} - 2^{x + 1} - 8 = 0.

Since $2^{2 x} = (2^{x})^{2}$ and $2^{x + 1} = 2 \cdot 2^{x}$ , setting $t = 2^{x}$ , with $t > 0$ :

t^{2} - 2 t - 8 = 0 ⟺ t = 4 or t = - 2

The solution $t = - 2$ must be discarded, because $t > 0$ .

2^{x} = 4 = 2^{2} ⟺ x = 2

Solution: $x = 2$ .

The term $2^{x + 1}$ is rewritten as $2 \cdot 2^{x}$ with the product rule for powers, so the whole equation is expressed in terms of $2^{x}$ and the substitution $t = 2^{x}$ gives a second-degree equation. Of the two solutions, $t = - 2$ is negative and therefore must be discarded, because $2^{x} > 0$ for every $x$ . What remains is $2^{x} = 4$ , which by the equality of powers with the same base gives $x = 2$ . The check confirms the result: for $x = 2$ we obtain $16 - 8 - 8 = 0$ .

Exercise 3

Solve the equation

3^{x} + 3^{1 - x} = 4.

Since $3^{1 - x} = \frac{3}{3 ^{x}}$ , setting $t = 3^{x}$ , with $t > 0$ :

t + \frac{3}{t} = 4 ⟺ t^{2} - 4 t + 3 = 0 ⟺ t = 1 or t = 3

3^{x} = 1 = 3^{0} ⟺ x = 0, 3^{x} = 3 ⟺ x = 1

Solutions: $x = 0$ and $x = 1$ .

The term $3^{1 - x}$ is rewritten as $\frac{3}{3 ^{x}}$ with the quotient rule for powers, so the substitution $t = 3^{x}$ gives the equation $t + \frac{3}{t} = 4$ . Since $t > 0$ , multiplying both sides by $t$ is legitimate and introduces no extraneous solutions; this gives the second-degree equation $t^{2} - 4 t + 3 = 0$ , with solutions $t = 1$ and $t = 3$ , both positive and therefore acceptable. The two equations with the same base give $x = 0$ and $x = 1$ ; the check confirms them, because $1 + 3 = 4$ and $3 + 1 = 4$ .