Exponential equations

This page is part of the section “Task 1 USE profile level. Elementary equations”. Here are solutions to exponential equations with neat integer roots. Each problem includes the condition, a step-by-step solution, and the answer.

Problems with solutions

Problem 1. Reducing to a common base

Solve the equation:

2^{x} = 8

Represent 8 as a power of two: 8 = 2³. Then the equation takes the form:

8 = 2^{3} \Rightarrow 2^{x} = 2^{3} \Rightarrow x = 3

Answer: $x = 3$ .

Problem 2. Power in the exponent

Solve the equation:

3^{x + 1} = 27

Represent 27 as a power of three: 27 = 3³. We get:

27 = 3^{3} \Rightarrow 3^{x + 1} = 3^{3}

Since the base is greater than 1, the exponents are equal:

x + 1 = 3 \Rightarrow x = 2

Answer: $x = 2$ .

Problem 3. Double exponent

Solve the equation:

5^{2 x - 1} = 125

Represent 125 as a power of five: 125 = 5³. We get:

125 = 5^{3} \Rightarrow 5^{2 x - 1} = 5^{3}

Equate the exponents and solve the linear equation:

2 x - 1 = 3 \Rightarrow 2 x = 4 \Rightarrow x = 2

Answer: $x = 2$ .

Problem 4. Substitution

Solve the equation:

4^{x} - 3 \cdot 2^{x} + 2 = 0

Note that 4^x = (2²)^x = (2^x)². This reduces the equation to a quadratic one in 2^x:

4^{x} = {(2^{2})}^{x} = {(2^{x})}^{2}

Introduce the substitution:

t = 2^{x}, t > 0

We obtain the quadratic equation:

t^{2} - 3 t + 2 = 0 \Rightarrow t_{1} = 1, t_{2} = 2

Return to the variable x:

2^{x} = 1 \Rightarrow x = 0; 2^{x} = 2 \Rightarrow x = 1

Answer: $x_{1} = 0$ , $x_{2} = 1$ .

Frequently asked questions

How to solve exponential equations from Task 1 of the USE?

The main technique is to reduce both sides to a common base. If this succeeds, the exponents are equated. If not, substitution (e.g. t = 2^x) or taking logarithms is used.

Why can the exponents be equated when reducing to a common base?

Because the exponential function a^x for a > 0 and a ≠ 1 is strictly monotone: it takes each of its values exactly once. Hence, from a^u = a^v it follows that u = v.

When is substitution needed?

When the equation contains powers with different exponents but the same base, for example 4^x − 3·2^x + 2 = 0. The substitution t = 2^x turns it into a quadratic equation in t.

Why must t > 0 be required after the substitution t = 2^x?

Because 2^x > 0 for any real x. Negative and zero values of t correspond to no x, so the roots t ≤ 0 are rejected.

Exponential equations

Problems with solutions

Problem 1. Reducing to a common base

Problem 2. Power in the exponent

Problem 3. Double exponent

Problem 4. Substitution

Frequently asked questions

See also