Class 9 Advanced Maths Chapter 2 Logarithms Ex 2.4 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 2 Logarithms Ex 2.4 to verify their answers.

Advanced Maths Class 9 Exercise 2.4 Solutions

Class 9 Advanced Maths Ex 2.4 Solutions

Question 1.
Solve for x.
(a) log₃ (2x – 5) = 2
Solution:
Convert to exponential form first using:
b^y = x ⇔ log_b (x) = y
Exponential form: 2x – 5 = 3² = 9
⇒ x = 7

(b) log₇ (3x) + log₇ 2 = log₇ 24
Solution:
Simplify to get one log term using Product Rule:
log_ax + log_a y = log_a(xy)
log₇ (3x) + log₇ 2 = log₇ 6x
⇒ log₇ 6x = log₇ 24
Since the bases are the same, we can set the arguments equal to each other:
⇒ 6x = 24 ⇒ x = 4

(c) log₅ (x + 3) – log₅ (x – 1) = 1
Solution:
Simplify to get one log term using Quotient Rule:
log_a x – log_a y = log_a\(\left(\frac{x}{y}\right)\)
⇒ log₅(x + 3) – log₅(x – 1) = log₅\(\frac{x+3}{x-1}\)
Now, convert to exponential form using:
b^y = x ⇔ log_b (x) = y
log₅\(\frac{x+3}{x-1}\) = 1
⇒ \(\frac{x+3}{x-1}\) = 5¹
Solve for x: ⇒ (x + 3) = 5(x – 1)
⇒ x = 2

(d) log₂(x² – 7) = 3
Solution:
Convert to exponential form using:
b^y = x ⇔ log_b (x) = y
log₂(x² – 7) = 3
⇒ x² – 7 = 2³ = 8
⇒ x² = 15
Taking the square root on both sides:
⇒ x = ±\(\sqrt{15}\)

Question 2.
Solve for x.
(a) log₂(x – 3) + log₂(x + 1) = 5
Solution:
Using Product Rule:
log₂ (x – 3) + log₂(x + 1) = log₂ (x – 3)(x + 1) = 5
Using b^y = x ⇔ log_b (x) = y,
we get: x² – 2x – 3 = 2⁵ = 3²
Now, solve the quadratic equation and find the value of x:
⇒ x² – 2x – 3 = 2⁵ = 3²
⇒ x² – 2x – 35 = 0
⇒ (x – 7)(x + 5) = 0
⇒ x = 7, -5
After plugging in the values of x into the given equation, for x = -5 equation is not defined.
Final answer is x = 7.

(b) 2 log₄ x = log₄ (5x – 4)
Solution:
Using Power Rule on the left side of the equation, we get:
log₄ x² = log₄ (5x – 4)
Now, set the arguments equal:
x² = 5x – 4
Rearrange and solve the quadratic equation:
x²– 5x + 4 = 0
⇒ x = 4,1
After plugging in the values of x into the given equation, for both values of x, the given equation is defined.
Final answer is x = 1, 4.

(c) log₅ (x + 2) + log₅ (x – 2) = 1
Solution:
On applying the multiplication rule, we get:
log₅ (x + 2)(x – 2) = 1
Using b^y = x ⇔ log_b (x) = y,
we get: (x + 2)(x – 2) = 5¹
Now, solve the quadratic equation and find the value of x: ⇒ x² – 4 = 5
⇒ x² = 9
⇒ x = +3
As x = -3, would make the argument (x – 2) negative, and logarithmic term undefined.
Final answer is x = +3.

(d) log₁₀(x – 2) + log₁₀(x + 1) = 1
Solution:
Using Product Rule on left side of the equation:
log₁₀ (x – 2)(x + 1) = 1
⇒ (x – 2)(x + 1) = 10¹
⇒ x² -x – 2 = 10
⇒ (x – 4)(x + 3) = 0
⇒ x = 4, -3
Value of x = -3 would make the arguments of logarithmic terms negative, hence undefined.
Final answer is x = 4.

Question 3.
Solve for x.
(a) log_x (3x + 10) = 2, where x > 0 and x ≠ 1
Solution:
Using Power Rule,
we get: (3x + 10) = x²
Now, solve the quadratic:
x² – 3x – 10 = 0
⇒ (x – 5)(x + 2) = 0
⇒ x = 5, -2
The problem states x > 0. Therefore, we must discard -2. Final Answer: x = 5.

(b) (log₃ x² – 4 log₃ x + 3 = 0
Solution:
The given equation is quadratic in form. To make it easier to see, let u = log₃ x.
Given equation becomes:
u² – 4u + 3 = 0
Solve the quadratic in u:
(u – 3)(u – 1) = 0
⇒ u = 3,1
Substitute the value of u to find x:
log₃ x = 3
⇒ x = 3³ = 2 7
log₃ x = 1
⇒ x = 3¹ = 3
Final value of x is 3, 27.

(c) (log₂x)²+ log₂x³ = 10
Solution:
Apply the Power Rule to the second term:
(log₂ x)² + 3 log₂x = 10
The given equation is quadratic in form. To make it easier to see, let u = log₂ x.
⇒ u² + 3u – 10 = 0
Solve the quadratic in u:
⇒ (u + 5 )(u – 2) = 0
⇒ u = -5, 2
Substitute back to find x:
log₂x = -5
⇒ x = 2^-5 = \(\frac{1}{32}\)
log₂ x = 2
⇒ x = 2² = 4
Final answer: x = 4, \(\frac{1}{32}\).

(d) x^log₁₀x = 1000x²
Solution:
When the variable is in the exponent, the best strategy is to take the log of both sides. Since the exponent uses base 10, we will take log₁₀ of both sides.
⇒ log₁₀x^log₁₀x = log₁₀ (1000x²)
⇒ (log₁₀ x )(log₁₀ x) = (log₁₀ 1000) + (log₁₀ x²)
⇒ (log₁₀ x)² = 3 + 2(log₁₀x)
Assume u = (log₁₀ x), and solve the quadratic in u:
⇒ u² = 3 + 2u
⇒ (u – 3)(u + 1) = 0
⇒ u = 3, -1
Substitute u back to find x:
⇒ log₁₀ x = 3
⇒ x = 10³ = 1000
⇒ log₁₀ x = -1
⇒ x = 10^-1 = 0.1
Final answer: x = 1000, 0.1.

Question 4.
Solve for x.
(a) log₃(x – 1) = log₃ (2x -1)
Solution:
Since both sides are logs with the same base, we can set the arguments equal:
⇒ x² – 1 = 2x – 1
⇒ x(x – 2) = 0
⇒ x = 0 or x = 2
If x = 0, the argument (x² – 1) becomes -1. Logs of negative
numbers are undefined.
Final Answer: x = 2

(b) log_x 5 – log_x 2 = log_x\(\sqrt{x}\)
Solution:
Use Quotient Rule on left side:
log_x\(\frac{5}{2}\) = log_x\(\sqrt{x}\)
Now, equate the arguments:
\(\frac{5}{2}\) = \(\sqrt{x}\)
Squaring both sides: \(\frac{25}{4}\) = x
⇒ x = 6.25

(c) log₂ x + \(\frac{1}{\log _x 2}\) = 4
Solution:
Use Reciprocal property term:(log_ab = \(\frac{1}{\log _b a}\)) on second term:
⇒ log₂ x + log₂ x = 4
⇒ 2 log₂ x = A
⇒ x = 2² = 4

(d) log₃ (3 + x) + log₃ (8 – x) – log₃ (9x – 8) = 2- log₃ 9
Solution:
Simplify the right side:
2 – log₃ 3² = 2 – 2log₃3 = 2 – 2 = 0
Combine terms on the left side using Product Rule and
Quotient Rule: log₃ \(\frac{(3+x)(8-x)}{(9 x-8)}\)
Now, equate both sides and use Power Rule:
\(\frac{(3+x)(8-x)}{(9 x-8)}\) = 3⁰ = 1
Rearrange and simplify:
⇒ (3 + x)(8 – x) = (9x – 8)
⇒ x² + 4x – 32 = 0
⇒ x = -8, 4
Value x = -8, will make the argument (3 + x) negative and hence the logarithmic term will be undefined.
Final answer: x = A.

(e) log₁₀ [log₂ (log₃ 9)] = 5x
Solution:
Solve this from the inside out:
Innermost term: log₃ 9 = log₃ 3² = 2
Next layer: log₂ 2 = 1
Outer term: log₁₀ 1 = 0
Solve for: 0 = 5x
⇒ x = 0

Question 5.
If x = log \(\frac{1}{2}\) + log \(\frac{2}{3}\) + log \(\frac{3}{4}\) + … + log \(\frac{99}{100}\), where all logs are to the base 10, then evaluate (x + 1) (x + 2) (x + 3) …(x + 99).
Solution:
Expand the expression of x using Quotient Rule:
⇒ x = log \(\frac{1}{2}\) + log \(\frac{2}{3}\) + log \(\frac{3}{4}\) + … + log \(\frac{99}{100}\)
= (log 1 – log 2) + (log 2 – log 3) + (log 3 – log 4) + … + log 99 – log 100
Notice the cancellation; all the terms cancel out except the first and last term:
⇒ x = log 1 – log 100
= log 1 – log 10² = 0 – 2 log 10
Since log is to the base of 10, that means log 10 = 1
⇒ x = -2(1) = -2
Now, substitute the value of x in the given expression:
⇒ (x + a)(x + 2)(x + 3)… (x + 99)
= (-2 + 1)(-2 + 2)(-2 + 3)… (-2 + 99)
Notice that the second term is (- 2 + 2) = 0. Hence, the whole product becomes zero.
⇒ (x + 1)(x + 2)(x + 3)… (x + 99) = 0