Logarithm questions with answers and concepts

Learn logarithm with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

117 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Logarithms are the inverse operation of exponentiation. While exponents ask 'what is the result of repeated multiplication?', logarithms ask 'how many times must we multiply to get this result?' The logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. Logarithms simplify complex calculations involving multiplication, division, and powers, and are essential in many areas of mathematics, science, and engineering.

Key formulas and rules

10 rules

Definition: $lo g_{b} (a) = x ⟺ b^{x} = a$ (where $b > 0, b \neq = 1$ , and $a > 0$ )

Log of 1 and base: $lo g_{b} (1) = 0$ and $lo g_{b} (b) = 1$

Product Rule: $lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$

Quotient Rule: $lo g_{b} (\frac{x}{y}) = lo g_{b} (x) - lo g_{b} (y)$

Power Rule: $lo g_{b} (x^{k}) = k lo g_{b} (x)$

Base Power Rule: $lo g_{b^{k}} (x) = \frac{1}{k} lo g_{b} (x)$

Change of Base: $lo g_{b} (a) = \frac{l o g _{c} ( a )}{l o g _{c} ( b )} = \frac{l n ( a )}{l n ( b )}$

Reciprocal Relation: $lo g_{b} (a) = \frac{1}{l o g _{a} ( b )}$

Inverse Property: $b^{l o g_{b} (a)} = a$

Multiplication identity: $lo g_{b} (a) \times lo g_{a} (b) = 1$

Key concepts

Understanding Logarithm Notation

In $lo g_{b} (a)$ , 'b' is called the base and 'a' is called the argument or antilogarithm. Common bases include 10 (common logarithm, written as log or lg) and e ~= 2.718 (natural logarithm, written as ln). The base must be positive and not equal to 1, and the argument must be positive. Logarithms are only defined for positive real numbers.

Fundamental Logarithm Laws

The three fundamental laws are: (1) Product Rule - log of a product equals sum of logs, (2) Quotient Rule - log of a quotient equals difference of logs, and (3) Power Rule - log of a power equals exponent times log of base. These laws allow us to break down complex expressions into simpler components and are the foundation of all logarithmic calculations.

Change of Base Formula

The change of base formula $lo g_{b} (a) = lo g_{c} (a) / lo g_{c} (b)$ is crucial when you need to evaluate logarithms with bases not available on standard calculators. Most calculators only compute log base 10 and log base e (ln). To find $lo g_{2} (8)$ , you would calculate $ln (8) / ln (2)$ or $lo g (8) / lo g (2)$ , both giving 3. This formula is also useful for comparing logarithms with different bases.

Characteristic and Mantissa

For common logarithms (base 10), any positive number can be written in scientific notation as m x $1 0^{n}$ where 1 <= m < 10. The logarithm is then n + log(m), where n is the characteristic (integer part) and log(m) is the mantissa (decimal part, always positive). The characteristic of log(N) is: (number of digits in N) - 1 for N > 1, and negative for 0 < N < 1.

Antilogarithms

The antilogarithm is the inverse operation of logarithm. If $lo g_{b} (x) = y$ , then antilog_b(y) = $x = b^{y}$ . To find the antilog, you raise the base to the power of the given logarithm value. For common logarithms, antilog(x) = $1 0^{x}$ . For natural logarithms, antilog(x) = $e^{x}$ . Antilogarithms are used to recover original numbers from their logarithmic values.

Solving Logarithmic Equations

To solve equations involving logarithms: (1) Use logarithm laws to simplify expressions, (2) Convert to exponential form when beneficial, (3) Always check that solutions satisfy the domain (argument > 0, base > 0, base != 1). For equations like $lo g_{b} (x) = lo g_{b} (y)$ , we get $x = y$ . For $lo g_{b} (x) = c$ , we get $x = b^{c}$ . Multiple logarithmic terms often require combining them first.

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Worked examples

Example 1

Find the value of $lo g_{2} (32) + lo g_{3} (27) - lo g_{5} (125)$ .

Solution

Step 1: Evaluate each term using power properties.

lo g_{2} (32) = lo g_{2} (2^{5}) = 5

lo g_{3} (27) = lo g_{3} (3^{3}) = 3

lo g_{5} (125) = lo g_{5} (5^{3}) = 3

Step 2: Combine terms.

5 + 3 - 3 = 5

Answer: 5.

Example 2

Given $lo g_{10} (2) = 0.3010$ and $lo g_{10} (3) = 0.4771$ , compute $lo g_{10} (72)$ .

Solution

Step 1: Factor 72 into prime powers.

72 = 8 \times 9 = 2^{3} \times 3^{2}

Step 2: Apply logarithm product and power rules.

lo g_{10} (72) = lo g_{10} (2^{3} \times 3^{2}) = 3 lo g_{10} (2) + 2 lo g_{10} (3)

Step 3: Substitute values.

lo g_{10} (72) = 3 (0.3010) + 2 (0.4771) = 0.9030 + 0.9542 = 1.8572

Answer: 1.8572.

Example 3

Solve for real $x$ : $lo g_{2} (x) + lo g_{2} (x - 2) = 3$ .

Solution

Step 1: Combine logarithms using the product rule.

lo g_{2} [x (x - 2)] = 3

x (x - 2) = 2^{3} = 8

x^{2} - 2 x - 8 = 0

Step 2: Factor the quadratic equation.

(x - 4) (x + 2) = 0 ⟹ x = 4 or x = - 2

Step 3: Domain restriction ( $x > 2$ ).
Since logarithms are only defined for positive arguments, $x = - 2$ is extraneous.

Answer: $x = 4$ .

Example 4

Solve $2^{x} = 5$ for $x$ given $lo g_{10} (2) \approx 0.3010$ .

Solution

Step 1: Take $lo g_{10}$ of both sides.

x lo g_{10} (2) = lo g_{10} (5)

Step 2: Use $lo g_{10} (5) = lo g_{10} (\frac{10}{2}) = 1 - lo g_{10} (2) = 1 - 0.3010 = 0.6990$ .

x = \frac{0.6990}{0.3010} \approx 2.322

Answer: Approximately 2.322.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

If $lo g_{10} (2) = 0.3010$ and $lo g_{10} (3) = 0.4771$ , find $lo g_{10} (15)$ .

1.1761
1.7781
0.7781
1.0791

Answer: A. 1.1761

Explanation

$lo g_{10} (15) = lo g_{10} (3 \times 5) = lo g_{10} (3) + lo g_{10} (5)$
We need $lo g_{10} (5) = lo g_{10} (10/2) = lo g_{10} (10) - lo g_{10} (2) = 1 - 0.3010 = 0.6990$
$lo g_{10} (15) = 0.4771 + 0.6990 = 1.1761$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

If $lo g_{10} (2) = 0.3010$ and $lo g_{10} (3) = 0.4771$ , find $lo g_{10} (12)$ .

1.5563
1.0791
1.2553
1.7781

Answer: B. 1.0791

Explanation

$lo g_{10} (12) = lo g_{10} (4 \times 3) = lo g_{10} (4) + lo g_{10} (3) = lo g_{10} (2^{2}) + lo g_{10} (3) = 2 \times lo g_{10} (2) + lo g_{10} (3) = 2 \times 0.3010 + 0.4771 = 0.6020 + 0.4771 = 1.0791$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

If $lo g (x + y) = lo g x + lo g y$ , then find x in terms of y.

y/(y-1)
y/(1-y)
(y-1)/y
(1-y)/y

Answer: A. y/(y-1)

Explanation

$lo g (x + y) = lo g x + lo g y = lo g (x y)$
$x + y = x y$
$x + y - x y = 0$
$x (1 - y) = - y$
$x = - y / (1 - y) = y / (y - 1)$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always verify the domain before solving: the argument must be positive and the base must be positive and not equal to 1.
Memorize common logarithm values: log_2(2)=1, log_2(4)=2, log_2(8)=3, log_2(16)=4, log_2(32)=5, log_3(9)=2, log_5(25)=2, log_10(100)=2.
When given log values like log(2) or log(3), look for ways to express the target number as products or powers of these known values.
For equations with logarithms of different bases, use the change of base formula to convert to a common base.
If log_b(x) = 0, then x = 1 (regardless of the base, as long as it's valid).
When simplifying complex logarithmic expressions, work from the inside out - simplify the arguments first before applying log rules.

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Next topic: Odd Man Out and Series

Logarithm questions with answers and concepts

Learn logarithm with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

117 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Key formulas and rules

10 rules

Definition: $lo g_{b} (a) = x ⟺ b^{x} = a$ (where $b > 0, b \neq = 1$ , and $a > 0$ )

Log of 1 and base: $lo g_{b} (1) = 0$ and $lo g_{b} (b) = 1$

Product Rule: $lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$

Quotient Rule: $lo g_{b} (\frac{x}{y}) = lo g_{b} (x) - lo g_{b} (y)$

Power Rule: $lo g_{b} (x^{k}) = k lo g_{b} (x)$

Base Power Rule: $lo g_{b^{k}} (x) = \frac{1}{k} lo g_{b} (x)$

Change of Base: $lo g_{b} (a) = \frac{l o g _{c} ( a )}{l o g _{c} ( b )} = \frac{l n ( a )}{l n ( b )}$

Reciprocal Relation: $lo g_{b} (a) = \frac{1}{l o g _{a} ( b )}$

Inverse Property: $b^{l o g_{b} (a)} = a$

Multiplication identity: $lo g_{b} (a) \times lo g_{a} (b) = 1$

Key concepts

Understanding Logarithm Notation

Fundamental Logarithm Laws

Change of Base Formula

Characteristic and Mantissa

Antilogarithms

Solving Logarithmic Equations

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

Find the value of $lo g_{2} (32) + lo g_{3} (27) - lo g_{5} (125)$ .

Solution

Step 1: Evaluate each term using power properties.

lo g_{2} (32) = lo g_{2} (2^{5}) = 5

lo g_{3} (27) = lo g_{3} (3^{3}) = 3

lo g_{5} (125) = lo g_{5} (5^{3}) = 3

Step 2: Combine terms.

5 + 3 - 3 = 5

Answer: 5.

Example 2

Given $lo g_{10} (2) = 0.3010$ and $lo g_{10} (3) = 0.4771$ , compute $lo g_{10} (72)$ .

Solution

Step 1: Factor 72 into prime powers.

72 = 8 \times 9 = 2^{3} \times 3^{2}

Step 2: Apply logarithm product and power rules.

lo g_{10} (72) = lo g_{10} (2^{3} \times 3^{2}) = 3 lo g_{10} (2) + 2 lo g_{10} (3)

Step 3: Substitute values.

lo g_{10} (72) = 3 (0.3010) + 2 (0.4771) = 0.9030 + 0.9542 = 1.8572

Answer: 1.8572.

Example 3

Solve for real $x$ : $lo g_{2} (x) + lo g_{2} (x - 2) = 3$ .

Solution

Step 1: Combine logarithms using the product rule.

lo g_{2} [x (x - 2)] = 3

x (x - 2) = 2^{3} = 8

x^{2} - 2 x - 8 = 0

Step 2: Factor the quadratic equation.

(x - 4) (x + 2) = 0 ⟹ x = 4 or x = - 2

Step 3: Domain restriction ( $x > 2$ ).
Since logarithms are only defined for positive arguments, $x = - 2$ is extraneous.

Answer: $x = 4$ .

Example 4

Solve $2^{x} = 5$ for $x$ given $lo g_{10} (2) \approx 0.3010$ .

Solution

Step 1: Take $lo g_{10}$ of both sides.

x lo g_{10} (2) = lo g_{10} (5)

Step 2: Use $lo g_{10} (5) = lo g_{10} (\frac{10}{2}) = 1 - lo g_{10} (2) = 1 - 0.3010 = 0.6990$ .

x = \frac{0.6990}{0.3010} \approx 2.322

Answer: Approximately 2.322.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

If $lo g_{10} (2) = 0.3010$ and $lo g_{10} (3) = 0.4771$ , find $lo g_{10} (15)$ .

1.1761
1.7781
0.7781
1.0791

Answer: A. 1.1761

Explanation

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

If $lo g_{10} (2) = 0.3010$ and $lo g_{10} (3) = 0.4771$ , find $lo g_{10} (12)$ .

1.5563
1.0791
1.2553
1.7781

Answer: B. 1.0791

Explanation

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

If $lo g (x + y) = lo g x + lo g y$ , then find x in terms of y.

y/(y-1)
y/(1-y)
(y-1)/y
(1-y)/y

Answer: A. y/(y-1)

Explanation

$lo g (x + y) = lo g x + lo g y = lo g (x y)$
$x + y = x y$
$x + y - x y = 0$
$x (1 - y) = - y$
$x = - y / (1 - y) = y / (y - 1)$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always verify the domain before solving: the argument must be positive and the base must be positive and not equal to 1.
Memorize common logarithm values: log_2(2)=1, log_2(4)=2, log_2(8)=3, log_2(16)=4, log_2(32)=5, log_3(9)=2, log_5(25)=2, log_10(100)=2.
When given log values like log(2) or log(3), look for ways to express the target number as products or powers of these known values.
For equations with logarithms of different bases, use the change of base formula to convert to a common base.
If log_b(x) = 0, then x = 1 (regardless of the base, as long as it's valid).
When simplifying complex logarithmic expressions, work from the inside out - simplify the arguments first before applying log rules.

Ready to test your understanding?

Work through 117 questions with explanations after each answer.

Start practice

Next topic: Odd Man Out and Series

Key formulas and rules

Key concepts

Understanding Logarithm Notation

Fundamental Logarithm Laws

Change of Base Formula

Characteristic and Mantissa

Antilogarithms

Solving Logarithmic Equations

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Understanding Logarithm Notation

Fundamental Logarithm Laws

Change of Base Formula

Characteristic and Mantissa

Antilogarithms

Solving Logarithmic Equations

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.