Class 9 Advanced Maths Chapter 2 Logarithms Ex 2.2 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 2 Logarithms Ex 2.2 to verify their answers.

Advanced Maths Class 9 Exercise 2.2 Solutions

Class 9 Advanced Maths Ex 2.2 Solutions

Question 1.
Express the following as a single logarithm.
(a) log 2 + 2 log 7
Solution:
log 2 + 2 log 7 = log 2 + log 7²
(Using Power Rule: log_aM^k = k log_aM)
= log (2 × 7²)
(Using Product Rule: log x + log y = log (xy)
= log (98)

(b) log₃ 8 + log₃ 5 – log₃ 4
Solution:
log₃ 8 + log₃5 – log₃ 4 = log₃(8 × 5) – log₃4
(Using Product Rule: log_ax + log_ay = log_a (xy))
= log₃ 40 – log₃ 4
(Using Quotient Rule: log_a x – log_a y = log_a \(\left(\frac{x}{y}\right)\))
= log₃ \(\left(\frac{40}{4}\right)\)
= log₃ 10

(c) log 5 + 2 log 3 – log 15
Solution:
log 5+2 log 3 – log 15 = log 5 + log 3² – log 15
(Using Power Rule: k log_a M = log_a M^k)
= log (5 × 3²) – log 15
(Using Product Rule: log_a x + log_a y = log_a (xy))
= log \(\left(\frac{5 \times 3^2}{15}\right)\) = log 3
(Using Quotient Rule: log_a x – log_a y = log_a\(\left(\frac{x}{y}\right)\))

(d) 2 + 2 log₃ 3
Solution:
2 + 2 log₅ 3 = 2(log₅ 5) + 2 log₅ 3
(Using property: log₁ a = 1) = log₅ 5² + log₅ 3²
(Using Power Rule: log_a M^k = k log_a M)
= log₅ (25 x 9) = log₅ 225
(Using Product Rule: log_a x + log_a y = log_a (xy))

(e) 3 – \(\left(\frac{1}{2}\right)\)log₃9
Solution:
3 – \(\left(\frac{1}{2}\right)\) log₃ 9 = 3(log₃ 3) – \(\left(\frac{1}{2}\right)\) log₃ 9
(Using property: log_a a = 1)
= log₃ 3³ – log₃ 9^1/2
(Using Power Rule: log_aMk = k log_a M)
= log₃ \(\left(\frac{27}{3}\right)\)
(Using Quotient Rule: log_a x – log_a y = log,= log_a \(\left(\frac{x}{y}\right)\))
= log₃ 9

(f) 1 + 2log₄3 – 31og₄4
Solution:
1 + 2 log₄ 3 – 3 log₄ 4 = 1(log₄ 4) + 2 log₄ 3 – 3 log₄ 4
(Using property: log_a a = 1)
= log₄4 + log₄ 3² – log₄ 4³
(Using Power Rule: log_a M^k = k log_a M)
= log₄ (4 × 9) – log₄ 64
(Using Product Rule: log_a x + log_a y = log_a (xy))
= log₄\(\frac{36}{64}\) = log₄\(\left(\frac{9}{16}\right)\)
(Using Power Rule: log_a M^k = k log_a M)

Question 2.
Find the exact value of
(a) log₁₁ 121
Solution:
log₁₁ 121 = log₁₁ 11²
= 2 × log₁₁ 11
(Using Power Rule: log_aM^k = k log_aM)
= 2 × 1 = 2
(Using property: log_a a = 1)

(b) log₇1
Solution:
For any base, b, the value of log_b 1 = 0, b > 0.
log₇ 1 = 0

(c) log₅ 625
Solution:
log₅ 625 = log₅ 54 = 4 log₅ 3
(Using Power Rule: log_aM^k = k log_a M)
= 4(1) = 4
(Using property: log_a a = 1)

(d) log₈ 8
Solution:
log₈ 8 = 1 (Using property: log_a a = 1)

(e) log 1000
Solution:
If no base is mentioned, we assume the logarithm has base 10.
log 1000 = log₁₀1000
=log₁₀ 10³ = 3 log₁₀ 10
(Using Power Rule: log_a M^k = k log_a M)
= 3(1) = 3
(Using property: log_a a = 1)

Question 3.
If log₂ 3 = p and log₂ 5 = q, write the following in terms of p and q:
(a) log₂15
Solution:
log₂15 = log₂ (3 × 5)
= log₂ 3 + log₂ 5
(Using Product Rule: log_a x + log_ay = log_a(xy))
= p + q

(b) log₂ 45
Solution:
log₂ 45 = log₂ (9 × 5)
= log₂ 9 + log₂ 5
(Using Product Rule: log_a x + log_a y = log_a (xy))
= log₂ 3² + log₂ 5
= 2 log₂ 3 + log₂ 5
(Using Power Rule: log_a M^k = k log_a M)
= 2p + q

(c) log₂ \(\left(\frac{5}{3}\right)\)
Solution:
log₂ \(\frac{5}{3}\) = log₂ 5 – log₂ 3
(Using Quotient Rule: log_a x – log_a y = log_a\(\left(\frac{x}{y}\right)\))
= q – p

(d) log₂ 10
Solution:
log₂ 10 = log₂ (5 × 2)
= log₂ 5 + log₂ 2
(Using Product Rule: log_a x + log_a y = log_a (xy))
= q + 1
(Using property: log_a a = 1)

Question 4.
Which of the following are true?
(a) If 2^x+1 = 3^x+1, then x + 1 = x + 2
Solution:
For equation 2^{(x + 1)} = 3^{(x + 2)} take log on both sides:
LHS: log 2^x+1 = (x + 1) log 2
RHS: log 3^x+1 = (x + 2) log 3
Clearly, RHS ≠ LHS
Hence, the statement is not true.

(b) log (x + 1) = log x
Solution:
Take LHS, i.e., log (x + 1), it cannot be further simplified.
Clearly, RHS ≠ LHS
Hence, the statement is not true.

(c) log_b b³ = 3
Solution:
Take LHS, i.e., log_b b³. It can be further simplified.
⇒ log_b b³ = 3 log_b b = 3 x 1 = 3
(Using log_a a = 1 property and Power Rule:
log_a M^k = k log_a M)
Clearly, RHS = LHS
Hence, the statement is true.

(d) Logarithm to base 1 is not defined.
Solution:
By mathematical definition, the logarithmic function log_bx requires the base b to be a positive real number not equal to 1 (b > 0 and b ≠ 1). Allowing a base of 1 violates the fundamental rules and functions of algebra.
Hence, the statement is true.

Question 5.
If log₂₀₂₆ x – log₂₀₂₆ y = a, log₂₀₂₆ y – log₂₀₂₆ z = b and log₂₀₂₆ z – log₂₀₂₆ x = c, then find the value of \(\left(\frac{x}{y}\right)^{b-c}\) × \(\left(\frac{y}{z}\right)^{c-a}\) × \(\left(\frac{z}{x}\right)^{a-b}\)
Solution:
Consider log₂₀₂₆ x – log₂₀₂₆ y = a
Using Quotient Rule, we get: log₂₀₂₆ \(\frac{x}{y}\) = a
Now, change the logarithmic form to the exponential form: \(\frac{x}{y}\) = (2026)^a
Similarly, we can get: \(\frac{y}{z}\) = (2026)^b and \(\frac{z}{x}\) = (2026)^c
Substitute the derived expressions into the required expression:
⇒ \(\left(\frac{x}{y}\right)^{b-c} \times\left(\frac{y}{z}\right)^{c-a} \times\left(\frac{z}{x}\right)^{a-b}\)
= (2026)^a(b-c) × (2026)^b(c-a) × (2026)^c(a-b)

= (2026)^{(ab-ac+bc-ba+ac-bc)}.
= (2026)°
= 1