Chapter Solution

Chapter-2

Exercise 2.1 — Solutions

1. Write an equivalent logarithmic statement for:

(a) 5³ = 125

a^b = c  ⟺  log_ac = b

Therefore,

log₅125 = 3

Answer: log₅125 = 3

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(b) 2⁵ = 32

Therefore,

log₂32 = 5

Answer: log₂32 = 5

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(c) 7⁻¹ = 1/7

Therefore,

log₇(1/7) = −1

Answer: log₇(1/7) = −1

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(d) 3^−1/2 = 1/√3

Therefore,

log₃(1/√3) = −1/2

Answer: log₃(1/√3) = −1/2

2. Write an equivalent exponential statement for:

(a) log₂16 = 4

log_ab = c  ⟺  a^c = b

Therefore,

2⁴ = 16

Answer: 2⁴ = 16

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(b) log₉81 = 2

Therefore,

9² = 81

Answer: 9² = 81

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(c) log₅√5 = 1/2

Therefore,

5^1/2 = √5

Answer: 5^1/2 = √5

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(d) log₂(1/2) = −1

Therefore,

2⁻¹ = 1/2

Answer: 2⁻¹ = 1/2

3. Find the value of:

(a) log₁₀1000

Since 10³ = 1000,
log₁₀1000 = 3

Answer: 3

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(b) log₆36

Since 6² = 36,
log₆36 = 2

Answer: 2

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(c) log₂64

Since 2⁶ = 64,
log₂64 = 6

Answer: 6

Exercise 2.2 — Solutions

1. Express the following as a single logarithm:

(a) log 2 + 2 log 7

Using the laws of logarithms:
2 log 7 = log 7² = log 49

Therefore,
log 2 + log 49 = log(2 × 49)

Answer: log 98

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(b) log₃ 8 + log₃ 5 − log₃ 4

Using:
log_am + log_an = log_a(mn)
and
log_am − log_an = log_a(m/n)

Therefore,
log₃(8 × 5/4) = log₃10

Answer: log₃ 10

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(c) log 5 + 2 log 3 − log 15

2 log 3 = log 3² = log 9

Therefore,
log 5 + log 9 − log 15 = log(5 × 9/15)

Answer: log 3

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(d) 2 + 2 log₅ 3

Write 2 as a logarithm to base 5:
2 = log₅5² = log₅25

Also,
2 log₅3 = log₅3² = log₅9

Therefore,
log₅25 + log₅9 = log₅(25 × 9)

Answer: log₅225

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(e) 3 − 1/2 log₃ 9

Write 3 as a logarithm to base 3:
3 = log₃27

Also,
1/2 log₃9 = log₃9^1/2 = log₃√9 = log₃3

Therefore,
log₃27 − log₃3 = log₃(27/3)

Answer: log₃9

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(f) 1 + 2 log₄ 3 − 3 log₄ 4

Write 1 as a logarithm to base 4:
1 = log₄4

Also,
2 log₄3 = log₄9

and
3 log₄4 = log₄4³ = log₄64

Therefore,
log₄4 + log₄9 − log₄64 = log₄(4 × 9/64)

Answer: log₄(9/16)

2. Find the exact value of:

(a) log₁₁121

Since 11² = 121,
log₁₁121 = 2

Answer: 2

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(b) log₇1

Since 7⁰ = 1,
log₇1 = 0

Answer: 0

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(c) log₅625

Since 5⁴ = 625,
log₅625 = 4

Answer: 4

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(d) log₈8

Since 8¹ = 8,
log₈8 = 1

Answer: 1

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(e) log 1000

Here the base is 10.
Since 10³ = 1000,
log 1000 = 3

Answer: 3

3. If log₂3 = p and log₂5 = q, write the following in terms of p and q:

(a) log₂15

15 = 3 × 5

Therefore,
log₂15 = log₂3 + log₂5 = p + q

Answer: p + q

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(b) log₂45

45 = 3² × 5

Therefore,
log₂45 = 2 log₂3 + log₂5 = 2p + q

Answer: 2p + q

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(c) log₂(5/3)

Using the quotient law:
log₂(5/3) = log₂5 − log₂3

Answer: q − p

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(d) log₂10

10 = 2 × 5

Therefore,
log₂10 = log₂2 + log₂5 = 1 + q

Answer: 1 + q

4. Which of the following are true?

(a) If 2^x+1 = 3^x+2, then x + 1 = x + 2

The bases 2 and 3 are different, so equality of the powers does not mean that their exponents are equal.

Answer: False

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(b) log (x + 1) = log x

For equal logarithms with the same base, their arguments must be equal. Thus this would require:
x + 1 = x

This is impossible.

Answer: False

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(c) log_bb³ = 3

Using the definition of logarithm:
log_bb³ = 3

This is true for a valid logarithm base, where b > 0 and b ≠ 1.

Answer: True

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(d) Logarithm to base 1 is not defined.

A logarithm base must be positive and cannot be equal to 1. Therefore, logarithm to base 1 is not defined.

Answer: True

5. If log₂₀₂₆x − log₂₀₂₆y = a, log₂₀₂₆y − log₂₀₂₆z = b and log₂₀₂₆z − log₂₀₂₆x = c, then find the value of:

(x/y)^b−c × (y/z)^c−a × (z/x)^a−b

From the given information,
log₂₀₂₆(x/y) = a

Therefore,
x/y = 2026^a

Similarly,
y/z = 2026^b

and
z/x = 2026^c

Substitute these values in the required expression:

(2026^a)^b−c × (2026^b)^c−a × (2026^c)^a−b

Using the law (a^m)ⁿ = a^mn:

= 2026^a(b−c) × 2026^b(c−a) × 2026^c(a−b)

= 2026^{a(b−c) + b(c−a) + c(a−b)}

Simplifying the exponent:
ab − ac + bc − ab + ac − bc = 0

Therefore,
2026⁰ = 1

Answer: 1

EXERCISE 2.3 — SOLUTIONS

1. Express the following in logarithmic form:

(a) 5⁴ = 625

Using a^b = c ⇒ log_a c = b:
log₅ 625 = 4

(b) 10⁻² = 0.01

log₁₀ 0.01 = −2

(c) 7⁰ = 1

log₇ 1 = 0

(d) 8¹ = 8

log₈ 8 = 1

2. Using the properties of logs, simplify: log₂ 16 + log₂ 4

Using the product rule:
log_bM + log_bN = log_b(MN)

log₂16 + log₂4 = log₂(16 × 4)
= log₂64
= 6

3. Evaluate:

(a) log₂ 256

Since 2⁸ = 256,
log₂256 = 8

(b) log₄ 16

Since 4² = 16,
log₄16 = 2

(c) log₅ 125

Since 5³ = 125,
log₅125 = 3

(d) log₁₀ 0.001

Since 10⁻³ = 0.001,
log₁₀0.001 = −3

4. If log₂ 7 = p and log₂ 3 = q, write in terms of p and q.

(a) log₂ 21

log₂21 = log₂(7 × 3)
= log₂7 + log₂3
= p + q

(b) log₂ 49

log₂49 = log₂(7²)
= 2log₂7
= 2p

(c) log₂(7/3)

log₂(7/3) = log₂7 − log₂3
= p − q

(d) log₂ 63

63 = 7 × 9 = 7 × 3²
log₂63 = log₂7 + log₂(3²)
= p + 2log₂3
= p + 2q

5. Real-world Application:

(a) If a star is 100 times brighter than another, find the magnitude difference.

Magnitude difference = 2.5 × log₁₀(brightness ratio)
= 2.5 × log₁₀100
= 2.5 × 2
= 5 magnitudes

(b) A solution has pH 3 and another has pH 6. How many times more acidic is the first solution?

Given:
pH = −log₁₀[H⁺]

For pH 3:
[H⁺] = 10⁻³

For pH 6:
[H⁺] = 10⁻⁶

Ratio of acidity = 10⁻³ / 10⁻⁶ = 10³ = 1000

Therefore, the first solution is 1000 times more acidic.

(c) A magnitude 9 earthquake occurs on the Richter Scale. How many times stronger is it than a magnitude 4 earthquake?

Difference in magnitude = 9 − 4 = 5

On the Richter scale, each increase of 1 corresponds to a 10-fold increase in amplitude.
Therefore:
10⁵ = 100,000

Hence, the magnitude 9 earthquake is 100,000 times stronger than the magnitude 4 earthquake.

6. True or False: Explain your reasoning.

(a) (1/3)log_bx = ∛x; x > 0

False.
The correct logarithmic property is:
(1/3)log_bx = log_b(x^1/3) = log_b(∛x)
It is not equal to ∛x itself.

(b) log₈e = 1/ln 8

True.
By the change-of-base formula:
log₈e = ln(e) / ln(8)
Since ln(e) = 1,
log₈e = 1/ln 8.

(c) Logarithm of a negative number is defined.

False for real logarithms.
A real logarithm is defined only when its argument is positive.
Therefore, log_b(−x) is not defined as a real number for x > 0.

(d) log_b(M + N) = log_bM + log_bN

False.
The addition rule does not apply to logarithms.
The correct product rule is:
log_b(MN) = log_bM + log_bN

(e) The base of the logarithm can be any real number.

False.
For a real logarithm, the base must satisfy:
b > 0 and b ≠ 1.
Therefore, the base cannot be any arbitrary real number.

7. Which is the greatest integer that is less than log₄9 + log₉28?

We need to find the integer immediately below:
log₄9 + log₉28.

First,
4¹ = 4 < 9 < 16 = 4²
Therefore:
1 < log₄9 < 2

Also,
9¹ = 9 < 28 < 81 = 9²
Therefore:
1 < log₉28 < 2

Hence:
2 < log₄9 + log₉28 < 4

To determine whether the sum is greater than 3, note that:
log₄9 = log₂3 > 1.5

Also, since 28 > 27 = 9^3/2,
log₉28 > 3/2.

Therefore:
log₄9 + log₉28 > 1.5 + 1.5 = 3.

The sum is also less than 4, so the greatest integer less than it is:
3

8. Evaluate the value of (x + 5y), where x = log_1.43(43/40) and y = (1/2)^log₂5.

First, find y:

y = (1/2)^log₂5
= (2⁻¹)^log₂5
= 2^−log₂5
= 1/2^log₂5
= 1/5

Now, find x:

x = log_1.43(43/40)
≈ 0.2021969

Therefore:
x + 5y = 0.2021969 + 5(1/5)
= 0.2021969 + 1
= 1.2021969

Hence, the answer is approximately 1.2022.

EXERCISE 2.4 — SOLUTIONS

1. Solve for x.

(a) log₃(2x − 5) = 2

By the definition of logarithm:
2x − 5 = 3²
2x − 5 = 9
2x = 14
x = 7

(b) log₇(3x) + log₇2 = log₇24

Using the product rule:
log₇(3x × 2) = log₇24
log₇(6x) = log₇24
6x = 24
x = 4

(c) log₅(x + 3) − log₅(x − 1) = 1

Using the quotient rule:
log₅ (x + 3)/(x − 1) = 1
(x + 3)/(x − 1) = 5
x + 3 = 5x − 5
8 = 4x
x = 2

(d) log₂(x² − 7) = 3

x² − 7 = 2³
x² − 7 = 8
x² = 15
x = ±√15

2. Solve for x.

(a) log₂(x − 3) + log₂(x + 1) = 5

log₂[(x − 3)(x + 1)] = 5
(x − 3)(x + 1) = 2⁵
x² − 2x − 3 = 32
x² − 2x − 35 = 0
(x − 7)(x + 5) = 0
x = 7 or x = −5

Both values satisfy the logarithm conditions: x > 3 or x < −1.

(b) 2log₄x = log₄(5x − 4)

2log₄x = log₄(x²)
Therefore:
x² = 5x − 4
x² − 5x + 4 = 0
(x − 1)(x − 4) = 0
x = 1 or x = 4

(c) log₅(x + 2) + log₅(x − 2) = 1

log₅[(x + 2)(x − 2)] = 1
x² − 4 = 5
x² = 9
x = ±3

Both values satisfy the logarithm conditions: x > 2 or x < −2.

(d) log₁₀(x − 2) + log₁₀(x + 1) = 1

log₁₀[(x − 2)(x + 1)] = 1
(x − 2)(x + 1) = 10
x² − x − 2 = 10
x² − x − 12 = 0
(x − 4)(x + 3) = 0
x = 4 or x = −3

3. Solve for x.

(a) log_x(3x + 10) = 2, where x > 0 and x ≠ 1

3x + 10 = x²
x² − 3x − 10 = 0
(x − 5)(x + 2) = 0
x = 5 or x = −2
Since x > 0, we reject x = −2.
x = 5

(b) (log₃x)² − 4log₃x + 3 = 0

Let y = log₃x.
Then:
y² − 4y + 3 = 0
(y − 1)(y − 3) = 0
y = 1 or y = 3

Therefore:
log₃x = 1 ⇒ x = 3
log₃x = 3 ⇒ x = 27
x = 3 or x = 27

(c) (log₂x)² + log₂x³ = 10

Using log₂x³ = 3log₂x:
(log₂x)² + 3log₂x = 10

Let y = log₂x.
y² + 3y − 10 = 0
(y + 5)(y − 2) = 0
y = −5 or y = 2

Therefore:
log₂x = −5 ⇒ x = 2⁻⁵ = 1/32
log₂x = 2 ⇒ x = 4
x = 1/32 or x = 4

(d) x^log₁₀x = 1000x²

Since x must be positive, divide both sides by x²:
x^{log₁₀x − 2} = 1000

Let t = log₁₀x. Then x = 10^t.
Therefore:
(10^t)^{t − 2} = 10³
10^{t(t − 2)} = 10³
t(t − 2) = 3
t² − 2t − 3 = 0
(t − 3)(t + 1) = 0

Thus t = 3 or t = −1.
If t = 3, x = 10³ = 1000.
If t = −1, x = 10⁻¹ = 1/10.
x = 1000 or x = 1/10

4. Solve for x.

(a) log₃(x² − 1) = log₃(2x − 1)

Since the bases are equal:
x² − 1 = 2x − 1
x² = 2x
x(x − 2) = 0
x = 0 or x = 2

For x = 0, x² − 1 = −1, so the logarithm is not defined.
Therefore:
x = 2

(b) log_x5 − log_x2 = log_x√x

Using the quotient rule:
log_x(5/2) = log_x(x^1/2)
Therefore:
5/2 = x^1/2
x = (5/2)²
x = 25/4

(c) log₂x + 1/log_x2 = 4

Using the reciprocal property:
1/log_x2 = log₂x

Therefore:
log₂x + log₂x = 4
2log₂x = 4
log₂x = 2
x = 4

(d) log₃(3 + x) + log₃(8 − x) − log₃(9x − 8) = 2 − log₃9

Since 2 = log₃9:
2 − log₃9 = 0

Combining the logarithms:
log₃ ((3 + x)(8 − x))/(9x − 8) = 0

Therefore:
((3 + x)(8 − x))/(9x − 8) = 1
(3 + x)(8 − x) = 9x − 8

24 + 5x − x² = 9x − 8
x² + 4x − 32 = 0
(x − 4)(x + 8) = 0
x = 4 or x = −8

The logarithm conditions require:
3 + x > 0,   8 − x > 0,   9x − 8 > 0.
Hence only x = 4 is valid.
x = 4

(e) log₁₀[log₂(log₃9)] = 5x

First:
log₃9 = 2
log₂2 = 1
log₁₀1 = 0

Therefore:
0 = 5x
x = 0

5. If x = log(1/2) + log(2/3) + log(3/4) + ... + log(99/100), where all logs are to the base 10, evaluate (x + 1)(x + 2)(x + 3) ... (x + 99).

Given:
x = log₁₀(1/2) + log₁₀(2/3) + log₁₀(3/4) + ... + log₁₀(99/100)

Using the property:
log a + log b = log(ab)

x = log₁₀ [(1/2)(2/3)(3/4) ... (99/100)]

The product telescopes:
(1/2)(2/3)(3/4) ... (99/100) = 1/100

Therefore:
x = log₁₀(1/100)
= log₁₀(10⁻²)
= −2

Now substitute x = −2:
(x + 1)(x + 2)(x + 3) ... (x + 99)
= (−1)(0)(1)(2) ... (97)

Since one of the factors is zero:
(x + 1)(x + 2)(x + 3) ... (x + 99) = 0