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

The main technique is to use the definition of the logarithm: log_a b = c is equivalent to a^c = b (for b > 0). If there are several logarithms, first combine them into one using properties (sum, difference, power), then apply the definition.

What is the domain and why check it?

The domain is the set of admissible values. For the logarithm log_a b we require b > 0, a > 0, a ≠ 1. After solving the equation, check that the found roots belong to the domain. Otherwise extraneous roots may appear.

Why do extraneous roots appear?

When passing from a logarithmic equation to an algebraic one (for example, by exponentiating or “removing” the logarithm), the domain may expand. Therefore checking the roots against the domain is mandatory.

How does lg differ from log and ln?

lg is the decimal logarithm (base 10), ln is the natural logarithm (base e ≈ 2.718), log with an indicated base is the logarithm to that base (e.g. log₂ is base 2).

Logarithmic equations

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

Problems with solutions

Problem 1. By the definition of the logarithm

Solve the equation:

{log}_{2} (x + 3) = 4

By the definition of the logarithm: log₂ A = 4 means 2⁴ = A. Hence:

x + 3 = 2^{4} \Rightarrow x + 3 = 16 \Rightarrow x = 13

Domain check: x + 3 = 16 > 0 — the root is valid.

Answer: $x = 13$ .

Problem 2. Logarithm with a coefficient

Solve the equation:

{log}_{3} (2 x - 1) = 2

By the definition of the logarithm: log₃ A = 2 means 3² = A. Hence:

2 x - 1 = 3^{2} \Rightarrow 2 x - 1 = 9 \Rightarrow 2 x = 10 \Rightarrow x = 5

Domain check: 2x − 1 = 9 > 0 — the root is valid.

Answer: $x = 5$ .

Problem 3. Decimal logarithm

Solve the equation:

lg (x^{2} - 3 x) = 1

lg is the decimal logarithm, i.e. log₁₀. By definition: x² − 3x = 10¹ = 10. We obtain a quadratic equation:

x^{2} - 3 x = 10^{1} \Rightarrow x^{2} - 3 x - 10 = 0

By Vieta’s theorem the roots are:

x_{1} = 5, x_{2} = - 2

Check the domain: the argument of the logarithm must be positive. For x = 5: 25 − 15 = 10 > 0 — valid. For x = −2: 4 + 6 = 10 > 0 — formally also valid.

Both roots fall within the domain, but in Task 1 of the USE usually only one answer is required — if the largest root is requested, the answer is 5.

Answer: $x = 5$ .

Problem 4. Sum of logarithms

Solve the equation:

{log}_{2} (x - 1) + {log}_{2} (x + 1) = 3

The sum of logarithms with the same base equals the logarithm of the product:

{log}_{2} [(x - 1) (x + 1)] = 3

By the definition of the logarithm:

(x - 1) (x + 1) = 2^{3} \Rightarrow x^{2} - 1 = 8 \Rightarrow x^{2} = 9

x = 3 или x = - 3

Check the domain: x − 1 > 0 and x + 1 > 0, i.e. x > 1. The root x = −3 is invalid (x − 1 = −4 < 0). The root x = 3 is valid.

Answer: $x = 3$ .

Logarithmic equations

Problems with solutions

Problem 1. By the definition of the logarithm

Problem 2. Logarithm with a coefficient

Problem 3. Decimal logarithm

Problem 4. Sum of logarithms

Frequently asked questions

See also