Why must the domain be found first?

A logarithm is defined only for a positive argument, and its base must be positive and not 1. Any answer outside the domain must be rejected.

How do you solve 4^(x−1) = 8?

Write both sides as powers of 2: 2^(2x−2) = 2³. The exponential function is one-to-one, so 2x − 2 = 3 and x = 5/2.

Why is x = −2 rejected in log x + log(x − 3) = 1?

The domain is x > 3, because both log x and log(x − 3) need positive arguments. Only x = 5 lies in it.

When does the inequality sign flip for logarithms?

When the base is between 0 and 1, the logarithmic function is strictly decreasing, so the sign reverses. For a base greater than 1 it stays the same.

Which board sets my paper if I study in Gujrat?

Gujrat district is listed under the Board of Intermediate and Secondary Education, Gujranwala. PECTAA writes the books. Your board sets the paper and the dates, so follow its notices.

Exponential and Logarithmic Equations and Inequalities: PECTAA Class 12 Maths, Punjab Board

An equation such as 2^(2x) = 3^(5x+3) cannot be solved by ordinary algebra, because the unknown sits in the exponent. Logarithms bring it down. The same tools handle inequalities, as long as you watch the domain and the direction of the sign.

Unit 1 of the PECTAA Mathematics 12 book is Graphical Representation of Functions. After transformations of graphs (section 1.9), section 1.10 solves exponential and logarithmic equations and inequalities. This note is for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Domain first. In these problems the variable is in an exponent or inside a logarithm. Exponential expressions are defined for every real number. A logarithm logₐ(u) needs u > 0, a > 0 and a ≠ 1. Find the domain before solving, then keep only answers inside it.

Tools for equations.

Same base: if a^m = a^n then m = n, because the exponential function is one-to-one.
Take logarithms of both sides when the bases differ, then use the power law: ln(aˣ) = x ln a.
Combine logarithms with the product law, log m + log n = log(mn), then change to exponential form: log₁₀ y = 1 means y = 10.

Tools for inequalities.

If the base is greater than 1 (such as 10 or e), the logarithmic function is strictly increasing: log u ≥ log v exactly when u ≥ v.
If the base is between 0 and 1, it is strictly decreasing, so the sign flips: log u ≥ log v exactly when u ≤ v.
When you divide by a logarithm, check its sign. log(5/2) is positive because 5/2 > 1.

Key points

Write the domain as an interval at the start.
Check every candidate answer against the domain.
Watch the base: below 1 reverses inequalities.
Final answers go in set or interval notation.

Worked examples

Example 1 (book Example 15). Solve 4^(x−1) = 8.

Domain: all real numbers.
4 = 2² and 8 = 2³, so 2^(2x−2) = 2³.
One-to-one: 2x − 2 = 3, so x = 5/2. Solution set {5/2}.

Example 2 (book Example 16, first part). Solve 2^(2x) = 3^(5x+3).

Domain: all real numbers.
Take ln: 2x ln 2 = (5x + 3) ln 3.
Collect x: (2 ln 2 − 5 ln 3)x = 3 ln 3, that is (ln 4 − ln 243)x = ln 27.
Quotient law: x ln(4/243) = ln 27, so x = ln 27 / ln(4/243) = log₍₄/₂₄₃₎ 27, about −0.80.

Example 3 (book Example 16, second part). Solve log x + log(x − 3) = 1.

Domain: x > 0 and x > 3, so ]3, ∞[.
Product law: log[x(x − 3)] = 1, so x(x − 3) = 10.
x² − 3x − 10 = 0 gives (x − 5)(x + 2) = 0, so x = 5 or x = −2.
Only 5 is in ]3, ∞[. Solution set {5}.

Example 4 (book Example 17). Solve logₓ(x² − 3x + 2) = 2.

Base: x > 0, x ≠ 1. Argument: x² − 3x + 2 > 0.
Exponential form: x² = x² − 3x + 2, so 3x − 2 = 0 and x = 2/3.
Check: the base 2/3 is allowed, and the argument is 4/9 > 0. Solution set {2/3}.

Example 5 (book Example 19). Solve 2^(x+1) > 5ˣ.

Write 2·2ˣ > 5ˣ, and divide by 2ˣ > 0: 2 > (5/2)ˣ.
Take log base 10 (increasing): log 2 > x log(5/2).
log(5/2) > 0, so x < log 2 / log(5/2) = log₍₅/₂₎ 2 ≈ 0.7565.
Solution set ]−∞, log₍₅/₂₎ 2[.

Example 6 (book Example 20). Solve log(2x − 1) ≥ log(5 − 3x).

Domain: 2x − 1 > 0 and 5 − 3x > 0, so 1/2 < x < 5/3.
Base 10 is increasing: 2x − 1 ≥ 5 − 3x, so 5x ≥ 6 and x ≥ 6/5.
With the domain: solution set [6/5, 5/3[.

Example 7 (book Example 21). Solve log₇/₉(x² − 3x + 2) ≥ log₇/₉(2x − 1).

Domain: x² − 3x + 2 > 0 gives x < 1 or x > 2; 2x − 1 > 0 gives x > 1/2. Combined: ]1/2, 1[ ∪ ]2, ∞[.
Base 7/9 is less than 1, so the sign flips: x² − 3x + 2 ≤ 2x − 1, that is x² − 5x + 3 ≤ 0.
Roots (5 ± √13)/2, so (5 − √13)/2 ≤ x ≤ (5 + √13)/2.
With the domain: [(5 − √13)/2, 1[ ∪ ]2, (5 + √13)/2].

Common mistakes in the exam

Skipping the domain and keeping x = −2 in Example 3.
Forgetting to flip the sign for a base between 0 and 1.
Writing log(m + n) = log m + log n. The product law is log(mn) = log m + log n.
Dividing by a logarithm without checking its sign.
Mixing open and closed brackets. A strict inequality or a domain edge gives an open end.

Exam technique

Use the layout Domain → Solve → Check → Solution set, as the book does.
Keep exact forms such as log₍₅/₂₎ 2 and add a decimal only if asked.
Draw a number line to combine the domain with the solved range.
Check your board's notice for the sections in your paper.

Practice questions

Solve 3^(x+1) = 27. Answer: x + 1 = 3, so x = 2.
State the domain of log(x − 4). Answer: x > 4, that is ]4, ∞[.
Solve log x + log 2 = 1. Answer: 2x = 10, so x = 5.
Solve log(x + 1) > log 3. Answer: Base 10 is increasing, so x + 1 > 3 and x > 2.
Solve log₁/₂ x > log₁/₂ 4 for x > 0. Answer: The base is below 1, so x < 4. Solution ]0, 4[.

Ready to practise this? Continue in 2nd Year Mathematics.