If 2160 = 2 a · 3 b · 5 c , find a , b and c . Hence calculate the value of 3 a × 2 - b × 5 - c .

Consider, 2160 = 2 a · 3 b · 5 c ⇒ 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2 a · 3 b · 5 c ⇒ 2 4 × 3 3 × 5 1 = 2 a · 3 b · 5 c Comparing the powers of 2 , 3 and 5 on both sides of above equation, we get a = 4 , b = 3 , c = 1 . Now, consider 3 a × 2 - b × 5 - c = 3 4 × 2 - 3 × 5 - 1 = 81 × 1 2 3 × 1 5 ∵ x - n = 1 x n = 81 × 1 8 × 1 5 = 81 40

If 1960 = 2 a · 5 b · 7 c , calculate the value of 2 - a · 7 b · 5 - c .

We have, 1960 = 2 a · 5 b · 7 c ⇒ 2 · 2 · 2 · 5 · 7 · 7 = 2 a · 5 b · 7 c ⇒ 2 3 · 5 1 · 7 2 = 2 a · 5 b · 7 c Comparing the powers of 2 , 5 and 7 on both sides of equation, we get a = 3 , b = 1 , c = 2 Consider, 2 - a · 7 b · 5 - c = 2 - 3 · 7 1 · 5 - 2 = 1 2 3 · 7 · 1 5 2 ∵ x - n = 1 x n = 1 8 · 7 · 1 25 = 7 200

Simplify: 8 3 a × 2 5 × 2 2 a 4 × 2 11 a × 2 - 2 a

We have, 8 3 a × 2 5 × 2 2 a 4 × 2 11 a × 2 - 2 a = 2 3 3 a × 2 5 × 2 2 a 2 2 × 2 11 a × 2 - 2 a = 2 3 × 3 a × 2 5 × 2 2 a 2 2 × 2 11 a × 2 - 2 a ∵ x n m = x n m = 2 9 a × 2 5 × 2 2 a 2 2 × 2 11 a × 2 - 2 a = 2 9 a + 5 + 2 a 2 2 + 11 a - 2 a ∵ x n · x m = x n + m = 2 11 a + 5 2 2 + 9 a = 2 11 a - 9 a + 5 - 2 ∵ x n x m = x n - m = 2 2 a + 3

Show that a m a - n m - n × a n a - l n - l × a l a - m l - m = 1

Consider, a m a - n m - n × a n a - l n - l × a l a - m l - m = a m - - n m - n × a n - - l n - l × a l - - m l - m ∵ x s x t = x s - t = a m + n m - n × a n + l n - l × a l + m l - m = a m - n m + n × a n - l n + l × a l - m l + m ∵ x s t = x s t = a m 2 - n 2 × a n 2 - l 2 × a l 2 - m 2 ∵ s + t s - t = s 2 - t 2 = a m 2 - n 2 + n 2 - l 2 + l 2 - m 2 ∵ x s · x t = x s + t = a 0 = 1 ∵ x 0 = 1 Hence, proved.

If a = x m + n · y l ; b = x n + l · y m and c = x l + m · y n prove that: a m - n · b n - l · c l - m = 1

Given: a = x m + n · y l , b = x n + l · y m and c = x l + m · y n . We have to prove that a m - n · b n - l · c l - m = 1 . Putting the values of a , b and c in a m - n b n - l c l - m = 1 , we get: = x m + n · y l m - n x n + l · y m n - l x l + m · y n l - m = x ( m + n ) ( m - n ) y l ( m - n ) x ( n + l ) ( n - l ) y m ( n - l ) x ( l + m ) ( l - m ) y n ( l - m ) ∵ ( A × B ) s = A s × B s ; A s t = A s × t = x m 2 - n 2 · x n 2 - l 2 · x l 2 - m 2 · y l m - l n · y m n - m l · y n l - n m ∵ ( A + B ) ( A - B ) = A 2 - B 2 = x m 2 - n 2 + n 2 - l 2 + l 2 - m 2 · y l m - l n + m n - m l + n l - n m ∵ A s × A t = A s + t = x 0 · y 0 = 1 × 1 ∵ A 0 = 1 = 1 . Hence, proved.

Simplify: x a x - b a 2 - a b + b 2 × x b x - c b 2 - b c + c 2 × x c x - a c 2 - c a + a 2

We have, x a x - b a 2 - a b + b 2 × x b x - c b 2 - b c + c 2 × x c x - a c 2 - c a + a 2 = x a + b a 2 - a b + b 2 × x b + c b 2 - b c + c 2 × x c + a c 2 - c a + a 2 ∵ A m A n = A m - n = x a + b a 2 - a b + b 2 × x b + c b 2 - b c + c 2 × x c + a c 2 - c a + a 2 ∵ A m n = A m n = x a 3 + b 3 × x b 3 + c 3 × x c 3 + a 3 ∵ A 3 + B 3 = A + B A 2 - A B + B 2 = x a 3 + b 3 + b 3 + c 3 + c 3 + a 3 ∵ A m × A n = A m + n = x 2 a 3 + 2 b 3 + 2 c 3 = x 2 a 3 + b 3 + c 3

E M B I B E

Concise Mathematics I.C.S.E - Class IX>Algebra>Chapter 7 - Indices [Exponents]>EXERCISE

R K Bansal Solutions for Exercise 1: EXERCISE

Author:R K Bansal

R K Bansal Mathematics Solutions for Exercise - R K Bansal Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Concise Mathematics I.C.S.E - Class IX solutions are prepared by Experienced Embibe Experts.

Questions from R K Bansal Solutions for Exercise 1: EXERCISE with Hints & Solutions

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 5

If $2160 = 2^{a} \cdot 3^{b} \cdot 5^{c}$ , find $a, b$ and $c$ . Hence calculate the value of $3^{a} \times 2^{- b} \times 5^{- c}$ .

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 6

If $1960 = 2^{a} \cdot 5^{b} \cdot 7^{c}$ , calculate the value of $2^{- a} \cdot 7^{b} \cdot 5^{- c}$ .

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 7

Simplify: $\frac{8^{3 a} \times 2^{5} \times 2^{2 a}}{4 \times 2^{11 a} \times 2^{- 2 a}}$

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 7

Simplify: $\frac{3 \times 27^{n + 1} + 9 \times 3^{3 n - 1}}{8 \times 3^{3 n} - 5 \times 27^{n}}$

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 8

Show that ${(\frac{a^{m}}{a^{- n}})}^{m - n} \times {(\frac{a^{n}}{a^{- l}})}^{n - l} \times {(\frac{a^{l}}{a^{- m}})}^{l - m} = 1$

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 9

If $a = x^{m + n} \cdot y^{l}; b = x^{n + l} \cdot y^{m}$ and $c = x^{l + m} \cdot y^{n}$ prove that: $a^{m - n} \cdot b^{n - l} \cdot c^{l - m} = 1$

EASY

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 10

Simplify: ${(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{a^{2} + a c + c^{2}}$

MEDIUM

9th ICSE

IMPORTANT

Concise Mathematics I.C.S.E - Class IX>Chapter 7 - Indices [Exponents]>EXERCISE>Q 10

Simplify: ${(\frac{x^{a}}{x^{- b}})}^{a^{2} - a b + b^{2}} \times {(\frac{x^{b}}{x^{- c}})}^{b^{2} - b c + c^{2}} \times {(\frac{x^{c}}{x^{- a}})}^{c^{2} - c a + a^{2}}$

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