Question 1

If log₅(x) = 2, then x equals:

Accepted Answer

We use the definition of logarithm: if logₐ(b) = c, then aᶜ = b.

Given: log₅(x) = 2

Step 1: Apply the definition of logarithm in exponential form.
log₅(x) = 2 means 5² = x

Step 2: Calculate 5².
x = 25

Verification: log₅(25) = log₅(5²) = 2 ✓

Question 2

Solve for x: log₂(x + 1) = log₂(5) + log₂(3)

Accepted Answer

We use the logarithm product rule: logₐ(m) + logₐ(n) = logₐ(mn).

Given: log₂(x + 1) = log₂(5) + log₂(3)

Step 1: Apply the product rule to the right side.
log₂(x + 1) = log₂(5 × 3)
log₂(x + 1) = log₂(15)

Step 2: Since the logarithm function is one-to-one (injective), if logₐ(m) = logₐ(n), then m = n.
x + 1 = 15

Step 3: Solve for x.
x = 14

Step 4: Check domain. For log₂(x + 1) to be defined, we need x + 1 > 0, so x > -1. Since x = 14 > -1, the solution is valid.

Verification: log₂(15) = log₂(5 × 3) = log₂(5) + log₂(3) ✓

Question 3

If log₂(x) + log₂(x - 2) = 3, then the value of x is:

Accepted Answer

We use the logarithm product rule: logₐ(m) + logₐ(n) = logₐ(mn).

Given: log₂(x) + log₂(x - 2) = 3

Step 1: Apply product rule.
log₂[x(x - 2)] = 3

Step 2: Convert to exponential form using the definition logₐ(b) = c ⟺ aᶜ = b.
x(x - 2) = 2³
x(x - 2) = 8

Step 3: Expand and rearrange.
x² - 2x = 8
x² - 2x - 8 = 0

Step 4: Factor the quadratic.
(x - 4)(x + 2) = 0
So x = 4 or x = -2

Step 5: Check domain restrictions.
For log₂(x) to be defined: x > 0
For log₂(x - 2) to be defined: x - 2 > 0, so x > 2

Therefore x must satisfy x > 2.
Only x = 4 satisfies this condition.

Verification: log₂(4) + log₂(2) = 2 + 1 = 3 ✓

Question 4

Simplify: log₃(27) + log₃(9) - log₃(81)

Accepted Answer

We use the power rule for logarithms: logₐ(bⁿ) = n·logₐ(b).

Step 1: Express each argument as a power of 3.
27 = 3³
9 = 3²
81 = 3⁴

Step 2: Apply the power rule.
log₃(3³) + log₃(3²) - log₃(3⁴)
= 3·log₃(3) + 2·log₃(3) - 4·log₃(3)

Step 3: Use the identity logₐ(a) = 1.
= 3(1) + 2(1) - 4(1)
= 3 + 2 - 4
= 1

Alternative approach using quotient rule:
log₃(27) + log₃(9) - log₃(81) = log₃[(27 × 9)/81]
= log₃[243/81]
= log₃(3)
= 1

Question 5

If log₅(x) = 2 and log₅(y) = -1, then log₅(x/y) equals:

Accepted Answer

We use the quotient rule for logarithms: logₐ(m/n) = logₐ(m) - logₐ(n).

Given:
log₅(x) = 2
log₅(y) = -1

Step 1: Apply the quotient rule.
log₅(x/y) = log₅(x) - log₅(y)

Step 2: Substitute the given values.
log₅(x/y) = 2 - (-1)
= 2 + 1
= 3

Verification (optional):
From log₅(x) = 2, we get x = 5² = 25
From log₅(y) = -1, we get y = 5⁻¹ = 1/5
So x/y = 25 ÷ (1/5) = 25 × 5 = 125 = 5³
Therefore log₅(125) = log₅(5³) = 3 ✓

Question 6

Solve for x: log₁₀(x + 1) = log₁₀(5) + log₁₀(2)

Accepted Answer

We use the product rule for logarithms: logₐ(m) + logₐ(n) = logₐ(mn).

Given: log₁₀(x + 1) = log₁₀(5) + log₁₀(2)

Step 1: Apply the product rule to the right side.
log₁₀(x + 1) = log₁₀(5 × 2)
log₁₀(x + 1) = log₁₀(10)

Step 2: Since the logarithm function is one-to-one, if logₐ(A) = logₐ(B), then A = B.
x + 1 = 10

Step 3: Solve for x.
x = 9

Step 4: Verify domain.
For log₁₀(x + 1) to be defined, we need x + 1 > 0, so x > -1.
Since x = 9 > -1, this is valid. ✓

Verification:
log₁₀(9 + 1) = log₁₀(10) = 1
log₁₀(5) + log₁₀(2) = log₁₀(10) = 1 ✓

Question 7

Simplify: log₃(27) + log₅(125) - log₂(8) = ?

Accepted Answer

Setup: Power rule of logarithms — testing whether students recognize perfect powers and apply log_a(aⁿ) = n correctly.

Solution:
Step 1: Express each argument as a power of the base.
27 = 3³
125 = 5³
8 = 2³

Step 2: Apply the power rule log_a(aⁿ) = n.
log₃(3³) = 3
log₅(5³) = 3
log₂(2³) = 3

Step 3: Substitute and compute.
3 + 3 - 3 = 3

Trap Explanation: Students may incorrectly compute log₃(27) as log₃(3 × 9) and try to use the product rule, or they may forget that log_a(aⁿ) = n and instead attempt to compute decimal approximations.

Shortcut: Recognize that 27 = 3³, 125 = 5³, and 8 = 2³ immediately. Then log_a(aⁿ) = n gives 3 + 3 - 3 = 3 in one line.

Question 8

If log₁₀(2) = 0.301 and log₁₀(3) = 0.477, then log₁₀(6) = ?

Accepted Answer

Setup: Product rule of logarithms log_a(mn) = log_a(m) + log_a(n) — testing whether students can decompose a number into prime factors and apply the rule.

Solution:
Step 1: Express 6 as a product of numbers whose logarithms are known.
6 = 2 × 3

Step 2: Apply the product rule.
log₁₀(6) = log₁₀(2 × 3) = log₁₀(2) + log₁₀(3)

Step 3: Substitute the given values.
log₁₀(6) = 0.301 + 0.477 = 0.778

Trap Explanation: Students may incorrectly apply the quotient rule (thinking 6 = 12/2) or the power rule (thinking 6 = 2³ or similar), leading to subtraction or multiplication of the given values instead of addition.

Shortcut: Recognize 6 = 2 × 3 immediately. Use product rule: log₁₀(6) = log₁₀(2) + log₁₀(3) = 0.301 + 0.477 = 0.778.

Question 9

Solve for x: log₂(log₂(x)) = 1

Accepted Answer

Setup: Composition of logarithmic functions — testing whether students can work backwards through nested logarithms using the definition log_a(y) = b ⟺ y = aᵇ.

Solution:
Step 1: Let y = log₂(x). Then the equation becomes:
log₂(y) = 1

Step 2: Convert from logarithmic to exponential form.
log₂(y) = 1 means y = 2¹ = 2

Step 3: Substitute back y = log₂(x).
log₂(x) = 2

Step 4: Convert to exponential form again.
log₂(x) = 2 means x = 2² = 4

Step 5: Verify domain constraints.
For log₂(log₂(x)) to be defined:
- x > 0 (so log₂(x) is defined)
- log₂(x) > 0 (so log₂(log₂(x)) is defined), which means x > 1

Since x = 4 > 1, it satisfies all constraints.

Verification: log₂(log₂(4)) = log₂(2) = 1 ✓

Trap Explanation: Students often make errors when working with nested functions. Common mistakes include: (1) applying the exponential conversion only once instead of twice, (2) forgetting domain constraints (log₂(x) must be positive, not just positive), or (3) confusing the order of operations.

Shortcut: Work from outside in. log₂(y) = 1 → y = 2. Then log₂(x) = 2 → x = 4. Check: x > 1 ✓

Question 10

If log_x(8) = 3/2, then x = ?

Accepted Answer

Setup: Change of base and exponential form of logarithms — testing whether students can convert log_x(8) = 3/2 to exponential form x^(3/2) = 8 and solve for x.

Solution:
Step 1: Convert from logarithmic to exponential form.
log_x(8) = 3/2 means x^(3/2) = 8

Step 2: Raise both sides to the power of 2/3 to isolate x.
(x^(3/2))^(2/3) = 8^(2/3)
x^((3/2)·(2/3)) = 8^(2/3)
x^1 = 8^(2/3)
x = 8^(2/3)

Step 3: Simplify 8^(2/3).
8^(2/3) = (8^(1/3))² = (∛8)² = 2² = 4

Alternatively: 8^(2/3) = (2³)^(2/3) = 2^(3·(2/3)) = 2² = 4

Step 4: Verify the solution.
log₄(8) = log₄(4^(3/2)) = 3/2 ✓
(Since 4^(3/2) = (2²)^(3/2) = 2³ = 8)

Trap Explanation: Students often struggle with fractional exponents. Common errors include: (1) incorrectly computing 8^(2/3) as 8² / 8³ or similar, (2) forgetting that (x^a)^b = x^(ab), or (3) computing the cube root and square incorrectly.

Shortcut: Recognize 8 = 2³. Then x^(3/2) = 2³ → x = 2² = 4 (by comparing exponents after expressing both sides as powers of 2).

Question 11

Simplify: log₃(27) + log₃(9) - log₃(3) = ?

Accepted Answer

We use logarithm properties: logₐ(aⁿ) = n, and logₐ(m) + logₐ(n) = logₐ(mn), and logₐ(m) - logₐ(n) = logₐ(m/n).

Given: log₃(27) + log₃(9) - log₃(3)

Step 1: Express each argument as a power of 3.
27 = 3³, so log₃(27) = 3
9 = 3², so log₃(9) = 2
3 = 3¹, so log₃(3) = 1

Step 2: Substitute the values.
3 + 2 - 1 = 4

Alternative approach using quotient and product rules:
log₃(27) + log₃(9) - log₃(3) = log₃(27 × 9 ÷ 3) = log₃(243/3) = log₃(81) = log₃(3⁴) = 4

Question 12

If log₁₀(a) = 2 and log₁₀(b) = 3, then log₁₀(a/b) equals:

Accepted Answer

We use the logarithm quotient rule: logₐ(m/n) = logₐ(m) - logₐ(n).

Given: log₁₀(a) = 2 and log₁₀(b) = 3

Step 1: Apply the quotient rule.
log₁₀(a/b) = log₁₀(a) - log₁₀(b)

Step 2: Substitute the given values.
log₁₀(a/b) = 2 - 3 = -1

Alternative verification:
From log₁₀(a) = 2, we get a = 10² = 100.
From log₁₀(b) = 3, we get b = 10³ = 1000.
So a/b = 100/1000 = 1/10 = 10⁻¹.
Therefore log₁₀(a/b) = log₁₀(10⁻¹) = -1 ✓

Question 13

If log₂(log₃(x)) = 1, then x equals:

Accepted Answer

We work from the outside logarithm inward, using the definition logₐ(b) = c ⟺ aᶜ = b.

Given: log₂(log₃(x)) = 1

Step 1: Convert the outer logarithm to exponential form.
log₂(log₃(x)) = 1 means 2¹ = log₃(x)
So log₃(x) = 2

Step 2: Convert the inner logarithm to exponential form.
log₃(x) = 2 means 3² = x
So x = 9

Step 3: Verify domain restrictions.
For log₂(log₃(x)) to be defined:
  - log₃(x) must be defined: x > 0
  - log₃(x) must be positive (since we take log₂ of it): log₃(x) > 0, which means x > 1

Since x = 9 > 1, both conditions are satisfied. ✓

Verification:
log₃(9) = log₃(3²) = 2
log₂(2) = 1 ✓

Question 14

If log₂(x + 1) + log₂(x - 1) = 3, then x equals:

Accepted Answer

We need to solve log₂(x + 1) + log₂(x - 1) = 3 using logarithm properties.

Step 1: Apply the product rule for logarithms: log_b(a) + log_b(c) = log_b(ac).
log₂(x + 1) + log₂(x - 1) = log₂[(x + 1)(x - 1)] = 3

Step 2: Convert from logarithmic to exponential form.
log₂[(x + 1)(x - 1)] = 3 means (x + 1)(x - 1) = 2³ = 8

Step 3: Expand the left side using difference of squares: (a + b)(a - b) = a² - b².
(x + 1)(x - 1) = x² - 1 = 8

Step 4: Solve the resulting quadratic equation.
x² - 1 = 8
x² = 9
x = ±3
So x = 3 or x = -3

Step 5: Apply domain constraints.
For log₂(x + 1) to be defined: x + 1 > 0 → x > -1
For log₂(x - 1) to be defined: x - 1 > 0 → x > 1
Both conditions must hold, so x > 1.
Since x = 3 > 1, it is valid.
Since x = -3 < 1, it is rejected.

Step 6: Verify the solution.
log₂(3 + 1) + log₂(3 - 1) = log₂(4) + log₂(2) = 2 + 1 = 3 ✓

Therefore, x = 3

Question 15

If log_x(64) = 2, then x equals:

Accepted Answer

We need to solve log_x(64) = 2 for the base x.

Step 1: Convert from logarithmic to exponential form.
log_x(64) = 2 means x² = 64

Step 2: Solve for x by taking square roots.
x = ±√64 = ±8
So x = 8 or x = -8

Step 3: Apply domain constraints for logarithm bases.
For a logarithm log_x(a) to be defined, the base x must satisfy:
  - x > 0 (base must be positive)
  - x ≠ 1 (base cannot equal 1)
Since x = -8 < 0, it is not a valid base.
Since x = 8 > 0 and x ≠ 1, it is valid.

Step 4: Verify the solution.
log₈(64) = log₈(8²) = 2 ✓

Therefore, x = 8

Question 16

If log₂(x² + 7x + 12) = 3, then the sum of all possible values of x is:

Accepted Answer

Setup: We need to solve a logarithmic equation by converting to exponential form, then solve the resulting quadratic. Step 1: Convert logarithmic to exponential form. log₂(x² + 7x + 12) = 3 ⟹ x² + 7x + 12 = 2³ = 8 Step 2: Rearrange to standard quadratic form. x² + 7x + 12 - 8 = 0 x² + 7x + 4 = 0 Step 3: Check domain constraint. For the logarithm to be defined: x² + 7x + 12 > 0 Factoring: (x + 3)(x + 4) > 0 This holds when x < -4 or x > -3 Step 4: Find roots of x² + 7x + 4 = 0 using Vieta's formulas. Sum of roots = -b/a = -7/1 = -7 Step 5: Verify both roots satisfy domain. Using quadratic formula: x = (-7 ± √(49 - 16))/2 = (-7 ± √33)/2 x₁ = (-7 + √33)/2 ≈ -0.628 (satisfies x > -3) ✓ x₂ = (-7 - √33)/2 ≈ -6.372 (satisfies x < -4) ✓ Both roots are valid. Sum = -7

Question 17

If log_a(b) · log_b(c) · log_c(a) = k, then k equals:

Accepted Answer

Setup: This tests the chain rule for logarithms and the relationship between logarithms with different bases.

Step 1: Apply the change of base formula.
Recall: log_x(y) = ln(y)/ln(x)

log_a(b) = ln(b)/ln(a)
log_b(c) = ln(c)/ln(b)
log_c(a) = ln(a)/ln(c)

Step 2: Multiply all three expressions.
log_a(b) · log_b(c) · log_c(a) = [ln(b)/ln(a)] · [ln(c)/ln(b)] · [ln(a)/ln(c)]

Step 3: Cancel common terms in numerator and denominator.
= [ln(b) · ln(c) · ln(a)] / [ln(a) · ln(b) · ln(c)]
= 1

Step 4: Verify with a specific example.
Let a = 2, b = 4, c = 8
log₂(4) = 2
log₄(8) = log₄(2³) = (3/2)log₄(2) = (3/2) · (1/2) = 3/4
log₈(2) = 1/3
Product: 2 · (3/4) · (1/3) = 6/12 = 1 ✓

Therefore, k = 1

Question 18

If log₂(log₃(log₅ x)) = 0, then the value of x is:

Accepted Answer

We work backwards from the outermost logarithm.

Given: log₂(log₃(log₅ x)) = 0

Step 1: Apply the definition of logarithm. If log₂(y) = 0, then y = 2⁰ = 1.
So: log₃(log₅ x) = 1

Step 2: Apply the definition again. If log₃(z) = 1, then z = 3¹ = 3.
So: log₅ x = 3

Step 3: Apply the definition once more. If log₅ x = 3, then x = 5³ = 125.

Verification:
- log₅(125) = log₅(5³) = 3 ✓
- log₃(3) = 1 ✓
- log₂(1) = 0 ✓

Therefore, x = 125.

Question 19

If log₁₀ 2 = 0.301 and log₁₀ 3 = 0.477, then log₁₀ 36 is equal to:

Accepted Answer

We express 36 in terms of the given logarithm values.

Step 1: Factorize 36.
36 = 4 × 9 = 2² × 3²

Step 2: Apply the logarithm product rule: log(ab) = log a + log b.
log₁₀ 36 = log₁₀(2² × 3²)
         = log₁₀(2²) + log₁₀(3²)

Step 3: Apply the logarithm power rule: log(aⁿ) = n·log a.
log₁₀(2²) = 2·log₁₀ 2 = 2 × 0.301 = 0.602
log₁₀(3²) = 2·log₁₀ 3 = 2 × 0.477 = 0.954

Step 4: Add the results.
log₁₀ 36 = 0.602 + 0.954 = 1.556

Verification: 10^1.556 ≈ 36 ✓

Question 20

The number of solutions to the equation log₂(x - 1) + log₂(x + 2) = 3 is:

Accepted Answer

Setup: Logarithmic equation requiring domain analysis, logarithm product rule, and quadratic solving.

Step 1: Identify domain constraints.
For log₂(x - 1) to be defined: x - 1 > 0 ⟹ x > 1
For log₂(x + 2) to be defined: x + 2 > 0 ⟹ x > -2
Combined domain: x > 1

Step 2: Apply the logarithm product rule.
log₂(x - 1) + log₂(x + 2) = 3
log₂[(x - 1)(x + 2)] = 3

Step 3: Convert to exponential form.
(x - 1)(x + 2) = 2³ = 8

Step 4: Expand and rearrange to standard quadratic form.
x² + 2x - x - 2 = 8
x² + x - 2 = 8
x² + x - 10 = 0

Step 5: Solve using the quadratic formula.
x = [-1 ± √(1 + 40)] / 2
x = [-1 ± √41] / 2

x₁ = (-1 + √41) / 2 ≈ (-1 + 6.403) / 2 ≈ 2.70
x₂ = (-1 - √41) / 2 ≈ (-1 - 6.403) / 2 ≈ -3.70

Step 6: Check against domain constraint (x > 1).
x₁ ≈ 2.70 > 1 ✓ Valid
x₂ ≈ -3.70 < 1 ✗ Invalid (violates domain)

Step 7: Verify the valid solution.
For x = (-1 + √41) / 2:
log₂((-1 + √41)/2 - 1) + log₂((-1 + √41)/2 + 2)
= log₂((-3 + √41)/2) + log₂((3 + √41)/2)
= log₂[(-3 + √41)(3 + √41) / 4]
= log₂[(41 - 9) / 4]
= log₂(32/4)
= log₂(8)
= 3 ✓

Therefore, there is exactly 1 solution.

AFCAT Logarithms

Logarithms— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Logarithms — Practice Questions

60-Second Revision — Logarithms