KSEEB Solutions for Class 8 Maths Chapter 10 Exponents Additional Questions

Students can Download Maths Chapter 10 Exponents Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 10 Exponents Additional Questions

Question 1.
Mark the correct option.
(a) The value of (3^m)ⁿ, for every pair of integers (m, n) is
A. 3^m+n
B. 3^mn
C. 3mⁿ
D. 3^m + 3ⁿ
Solution:
(B) 3^mn

(b) If x, y, 2x + \(\frac{\mathbf{y}}{2}\) are non-zero real numbers then \(\left(2 x+\frac{y}{2}\right)^{-1}\left[(2 x)^{-1}+\left(\frac{y}{2}\right)^{-1}\right]\) equals.
A. 1
B. x . y^-1
C. x^-1 . y
D. x^-1 . y^-1
Solution:
(D) x^-1 . y^-1

(c) If 2^x – 2^x-2 = 192 the value of x is
A. 5
B. 6
C. 7
D. 8
Solution:
(D) 8

(d) The number\(\left[6^{\left(6^{6}\right)}\right]^{\frac{1}{6}}\) is equal to
A. 6⁶
B. 6⁶ – 1
C. 6(6⁵)
D. \(6^{6^{5}}\)
Solution:
(C) 6(6⁵)

(e) The number of pairs of positive integers (m, n) such that mⁿ = 25 is
A. 0
B. 1
C. 2
D. More than 2
Solution:
(B) 1

Question 2.
The diameter of the sun is 1.4 x 109 meters and that of the earth is about 1.2678 x 10⁷ meteres. Find the approximate ratio of the diameter of the sun to that of the earth.
Solution:

= 1.10427 x 10^9-7
= 1.10427 x 10²
= 1.10427 x 100 = 110 . 427
∴ The ratio is 110 . 427 : 1

Question 3.
Find the value of each of the following expressions
(a) (-0.75)³ + (0.3)^-3 – \(\left(-\frac{3}{2}\right)^{-3}\)
Solution:

(b)
Solution:

(c)
Solution:

Question 4.
How many digits are there in the number 2³ x 5⁵ x 20⁵?
Solution:
2³ x 5⁵ x 20⁵
= 8 x 625 x 3200000 = 5000 x 32
= 16000000000
∴ The number has 11 digits.

Question 5.
If a7 = 3 find the of

Solution:

Question 6.
If 2^m. a² = 2⁸ where a, m are positive integers find all possible values of a + m.
Solution:
2^m . a² = 2⁸
2^m . a² = 2⁶ . 2²
∴m = 6 and a = 2
a + m = 6 + 2 = 8

Question 7.
Suppose 3^k x b² = 6⁴ for some positive integers k and b. Find all possible values of k + b.
Solution:
3^k x b² = 6⁴
= (3 x 2)⁴
= 3⁴ x 2⁴ = 3⁴ x 2^2×2 = 34 x (2²)²
= 3k x b² = 3⁴ x 4²
∴ k = 4 and b = 4
k + b = 4 + 4 = 8

Question 8.
Find the value of \(\frac{(625)^{6.25} \times(25)^{2.60}}{(625)^{7.25} \times(5)^{1.20}}\)
Solution:

Question 9.
A person had some rupees which is a power of 5. He gave a part of it to his friend which is also a power of 5. He was left with Rs. 500. How much did money he have?
Solution:
The powers of 5 are 5¹ = 5, 5² = 25, 5³
= 125, 5⁴ = 625,
5⁵ = 3125 etc.
We find that 625 – 125 = 500
∴ The money with him was Rs. 625.