Class 9 Advanced Maths Chapter 2 Logarithms Ex 2.3 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 2 Logarithms Ex 2.3 to verify their answers.

Advanced Maths Class 9 Exercise 2.3 Solutions

Class 9 Advanced Maths Ex 2.3 Solutions

Question 1.
Express the following in logarithmic form:
(a) 5⁴ = 625
Solution:
Using b^y = x ⇔ log_b (x) = y
Here, b = 5, y = 4, x = 625
Logarithmic form: log⁵ 625 = 4.

(b) 10^-2 = 0.01
Solution:
Using b^y = x ⇔ log_b (x) = y
Here, b = 10, y = – 2, x = 0.01
Logarithmic form: log₁₀ 0.01 = -2.

(c) 7° = 1
Solution:
Using b^y = x ⇔ log_b (x) = y
Here, b = 7, y = 0, x = 1
Logarithmic form: log₇ 1 = 0

(d) 8¹ = 8
Solution:
Using b^y = x ⇔ log_b (x) = y
Here, b = 8, y = 1, x = 8
Logarithmic form: log₇ 8 = 1

Question 2.
Using the properties of logs, simplify: log₂16 + log₂ 4
Solution:
log₂ 16 + log₂ 4 = log₂ (16 x 4)
(Using Product Rule: log_a x + log_a y = log_a (xy))
= log₂ 2⁶ = 6 log₂ 2
(Using property: log_a a = 1)
= 6(1) = 6

Question 3.
Evaluate:
(a) log₂ 256
Solution:
log₂ 256 = log₂ 2⁸ = 8 log₂ 2
(Using Power Rule: log_a M^k = k log_a M)
= 8
(Using property: log_a a = 1)

(b) log₄ 16
Solution:
log₄16 = log₄ 4² = 2 log₄ 4
(Using Power Rule: log_a M^k = k log_a M)
= 2
(Using property: log_a a = 1)

(c) log₅ 125
Solution:
log₅125 = log₅ 5³ = 3 log₅ 5
(Using Power Rule: log_a M^k = k log_a M)
= 3
(Using property: log_a a = 1)

(d) log₁₀ 0.001
Solution:
log₁₀0.001 = log₁₀10^-3 = -3 log₁₀10
(Using Power Rule: log_a M^k = k log_a M)
= -3
(Using property: log_a a = 1)

Question 4.
If log₂ 7 = p and log₂ 3 = q, write in terms of p and q.
(a) log₂ 21
Solution:
log₂ 21 = log₂ (7 × 3)
= log₂ 7 + log₂ 3
(UsingProduct Rule: log_a x + log_a y = log_a (xy))
= p + q

(b) log₂ 49
Solution:
log₂ 49 = log₂ 7² = 2 log₂ 7 = 2p
(Using Power Rule: log_a M^k = k log_a M)

(c) log₂\(\left(\frac{7}{3}\right)\)
Solution:
log₂\(\left(\frac{7}{3}\right)\) = log₂7 – log₂3
= p – q
(Using Quotient Rule: log_a x – log_a y = log₂\(\left(\frac{x}{y}\right)\))

(d) log₂ 63
Solution:
log₂ 63 = log₂ (7 × 9)
= log₂ 7 + log₂ 3²
(Using Product Rule: log_a x + log_a y = log_a (xy))
= log₂ 7 + 2 log₂ 3
(Using Power Rule: log_a M^k = k log_a M)
= P + 2q

Question 5.
Real-world Application:
(a) If a star is 100 times brighter than another, and the difference in their magnitudes is given by 2.5 × log₁₀ (“brightness ratio”), find the magnitude difference.
Solution:
Brightness Ratio = 100 Difference in their magnitudes,
∆m = 2.5 × log₁₀ (brightness ratio) = 2.5 × log₁₀ 100
∆m = 2.5 × log₁₀ 102 = 2.5 × (2 log₁₀ 10) = 2.5 × 2
∆m = 5
The magnitude difference is 5.

(b) A solution has pH 3 and another has pH 6. How many times more acidic is the first solution?
(pH = -log₁₀ [H⁺])
Solution:
Formula (pH = – log¹⁰ [H⁺]) can be rearranged to get the concentration of hydrogen ions as [H⁺] = 10^=pH.
For pH 3: [H⁺]₁ = 10^-3
For pH 6: [H⁺]₂ = 10^-6
Ratio: \(\frac{\left[\mathrm{H}^{+}\right]_1}{\left[\mathrm{H}^{+}\right]_2}=\frac{10^{-3}}{10^{-6}}\) = 1000
This shows that the solution with pH 3 is 1000 times more acidic than the solution with pH 6. This is because a 3-unit difference on a base-10 logarithmic scale corresponds to a factor of 10³.

(c) A magnitude 9 earthquake occurs on the Richter Scale. How many times stronger is it than a magnitude 4 earthquake?
Solution:
The Richter Scale is base-10 logarithmic. This means each increase of 1 on the scale represents a 10-fold increase in the measured amplitude of the earthquake.
Difference in magnitude, ∆m = 9 – 4 = 5
Ratio of amplitudes is 10^∆m = 10⁵
Hence, magnitude 9 earthquake is 100,000 times stronger than the earthquake of magnitude 4.

Question 6.
True or False: (Explain your reasoning)
(a) \(\frac{1}{3}\)log_bx =\(\sqrt[3]{x}\); x > 0
Solution:
False. Using the Power Rule, \(\frac{1}{3}\)log_bx = log_b x^1/3

(b) log₈ e = \(\frac{1}{\ln 8}\)
Solution:
True. This uses the Change of Base Formula:
log₈ e = \(\frac{\ln e}{\ln 8}=\frac{1}{\ln 8}\)

(c) Logarithm of a negative number is defined.
Solution:
False. For log_b x, x > 0. Value of x cannot be a negative number.

(d) log_b (M + N) = log_b M + log_b N
Solution:
False, as using Product Rule: log_a x + log_a y = log_a (xy)

(e) The base of the logarithm can be any real number.
Solution:
False. By definition, the base must be positive (b > 0) and not equal to 1.

Question 7.
Which is the greatest integer that is less than the number log₄ 9 + log₉ 28? (Do not use calculator)
Solution:
We need to find the greatest integer less than log₄ 9 + log₉ 28.
First, estimate each logarithm without a calculator.
Here, 4¹ = 4, 4² = 16 and 9 lies between 4 and 16
⇒ 1 < log₄ 9 < 2.
Also, 9¹ = 9, 9² = 81. And 28 lies between 9 and 81
⇒ 1 < log₉ 28 < 2.
Therefore, adding the two: 2 < log₄ 9 + log₉ 28 < 4.
More precisely,
4^{\(\frac{3}{2}\)} = 8 < 9 < 16 = 4²
⇒ \(\frac{3}{2}\) < log₄ 9 < 2 …(i)
9^{\(\frac{3}{2}\)} = 27 < 28 < 81 = 9²
⇒ \(\frac{3}{2}\) < log₉ 28 < 2 …(ii)
Adding Eqs (i) and (ii), we get,
3 < log₄ 9 + log₉ 28 < 4
Hence, the greatest integer less than the number log₄ 9 + log₉ 28 is 3.

Question 8.
Evaluate the value of (x + 5y), where x = \(\log _{1.4 \overline{3}}\left(\frac{43}{30}\right)\) and y = \(\left(\frac{1}{2}\right)^{\log _2 5}\).
Solution: Consider y first:
⇒ y = \(\left(\frac{1}{2}\right)^{\log _2 5}\) = (2^-1)^log₂2
= (2)^-log₂2 (Using: (a^-1)^b = a^-b)
⇒ y = 2^log₂(5-1)
(Using: Power Rule: -1.log_b a = log_ba^-1)
⇒ log₂y = log₂(5^-1)
(Using b^a = x ⇔ log_b (x) = a
By comparing arguments, we get, y = \(\frac{1}{5}\)
Now, consider x:
x = \(\log _{1.4 \overline{3}}\left(\frac{43}{30}\right)\), value of \(\frac{43}{30}=1.43333 \ldots=1.4 \overline{3}\)
⇒ x = \(\log _{1.4 \overline{3}}(1.4 \overline{3})\) = 1
Value of the expression: x + 5y = 1 + 5\(\left(\frac{1}{5}\right)\) = 2.