= log₅ 20 = log₅ 4 + log₅ 5

= log₅ 2² + 1 = 2log₅ 2 + 1

= log₃ 32 + log₇ 72

= 2log₃ 3 + 2log₇ 7

= 4

= log₄ (20/5) = log₄ 4 = 1

= 2 log₂ 2³ - 3log₆ 6³

= 6log₂ 2 - 9log₆6 = 6 - 9 = -3

= log₅ 50 + log₅ 10 - log₅ 4

= log₅ (2×5²) + log₅ 2 + log₅ 5 - log₅ 2²

= log₅ 2 + 2log₅ 5 + log₅ 2 + log₅ 5 - 2log₅ 2

= 3log₅5 = 3

= log₄ (4×5) + (log₄ (4²×2) - log₄ (2×5))

= log₄ 4 + log₄ 5 + 2log₄ 4 + log₄ 2

- log₄ 2 - log₄ 5

= 3

= log₈ 2⁵ + log₈ 2

=log₈ 64 = log₈ 2 = 2

= log₁₀ (125×32÷4) = log₁₀ 1000

= log₁₀ 10³ = 3

log(x + 1)(3 - x).

log (16x²)

log₅ (8x ÷ 2x) = log₅ 4.

log₁₀ (x²y³÷xy) = log₁₀ xy²

(= log₁₀ x + 2log₁₀y)

log₅ 200 - 3log₅ 2

= log₅ 200 - log₅ 2³

= log₅ (200/8)

= log₅ 25 = log₅ 5²

= 2 log₅ 5 = 2

(i) log₅ 25 = log₅ 5² = 2 log₅ 5 = 2.

(ii) Let x = log₅ 25.

5^x = 25 = 5²

x = 2

As the indices are added together, we can form two terms which are multiplied together:

e^ln2 × e^ln3 = 2 × 3 = 6.

As indices are subtracted from each other, we can form two terms which are divided:

.

LHS = log_a a² - log_a 75

= 2log_a a - (log_a25 + log_a 3)

= 2 - log_a 5² - log_a 3

= 2 - 2y - x

log_e m² = 2log_e m.

But if m = eⁿ, then log_e m = n

Hence log_e (m²) = 2n

3log₂ 8 = 3(log₂2³)

= 3(3log₂2)

= 3 × 3 = 9

log₁₀ (20A÷2A) = log₁₀10 = 1

Answer: (b) a = log_b 12.

Because:

Answer: (d) log₂ (16x).

Because:

4 + log₂ x = 4log₂ 2 + log₂ x

= log₂ (16 + log₂ x)

= log₂ (16x)

27. Given that

2log₃ (x²y) = 3 + log₃ x - log₃ y,

express y in terms of x.

log_a (21a) = log_a (3×7×a)

= log_a 3 + log_a 7 + log_a a

= 1.6 + 2.4 + 1 = 5

(a) log₂ 0.6;

(b) log₂ 60.

(a) log₂ 0.6 = log₂ 3/5

= log₂ 3 - log₂ 5

= 1.58 - 2.32 = - 0.74.

(b) log₂ 60 = log₂ (3×5×4)

= log₂3 + log₂ 5 +log₂ 2²

= 1.58 + 2.32 + 2 = 5.9

log_a 3 = 0.6826, find the value of log_a 24.

log_a 24 = log_a (8 × 3)

= 3log_a 2 + log_a 3

= 3×0.4307 + 0.6826

= 1.9747

(i) log_m (pq);

(ii) ;

(iii) pq² in terms of m.

(i) log_m (pq) = log_mp + log_mq

= 1.75 + 2.25 = 4.0

(ii) = log_mp - log_mq

= 1.75 - 2.25 = -0.5

(iii) Let y = pq²

log_m y = log_m(pq²)

= log_mp + 2log_mq

= 1.75 + 2×2.25 = 6.25

So y = m^6.25

log₃ 9 + log₃ x^½ - log₃ y

= 2log₃ 3 + ½log₃ x + log₃ y

= 2 + ½a + b

34. If log_a (xy³) = 1 and log_a (x²y) = 1, what is the value of log_a (xy)?

log_e x² = log_e (3x + 10)

Equating the terms after log_e

x² - 3x - 10 = 0

(x - 5)(x + 3) = 0

x = 5 or x = -3

But x cannot equal -3 as then the first term would be undefined.

So x = 5

log₅ 3 = 2log₅ 6 + log₅ x

log₅ 3 = log₅ 6² + log₅ x

log₅ 3 = log₅ (36x)

36x = 3

x = ¹/₁₂

(i) 2 log₁₀ x + 3 = 5log₁₀ x

3log₁₀ x = 3

log₁₀ x = 1

x = 10

OR

(ii) 2 log₁₀ x + 3log₁₀10 = 5log₁₀ x

x²× 10³ = x⁵

x³ = 1000

x = 10

log₂ (x - 1) = 5 - log₂ (x + 3).

log₂ (x - 1) + log₂ (x + 3)= 5

(x - 1)(x + 3) = 2⁵

x² + 2x - 35 = 0

(x + 7)(x - 5) = 0

x = -7 or x = 5

x cannot equal 7 as then both log terms would be negative.

x = 5

log₃ (2x - 1)(x - 4) = 2

2x² -9x + 4 = 9

2x² - 9x - 5 = 0

(2x + 1)(x - 5) = 0

x = -½ or x = 5

But x = -½ makes both log terms negative.

so x = 5

ln (x + 12) = ln x²

Equating the ln terms

x² - x - 12 = 0

(x - 4)(x + 3) = 0

x = 4 or -3

As x = -3 cannot be used in the RH term

x = 4

42.