# BSE Odisha 7th Class Maths Solutions Chapter 4 ଘାତାଙ୍କ ଓ ଘାତରାଶି Ex 4.2

Odisha State Board BSE Odisha 7th Class Maths Solutions Chapter 4 ଘାତାଙ୍କ ଓ ଘାତରାଶି Ex 4.2 Textbook Exercise Questions and Answers.

## BSE Odisha Class 7 Maths Solutions Chapter 4 ଘାତାଙ୍କ ଓ ଘାତରାଶି Ex 4.2

Question 1.
ଘାତାଙ୍କୀୟ ନିୟମ ବ୍ୟବହାର କରି ଏକ ଘାତରାଶିରେ ପରିଣତ କର ।

(କ) 23 × 24 × 25
ସମାଧାନ:
23 × 24 × 25 = 23+4+5 = 212

(ଖ) 615 ÷ 612
ସମାଧାନ:
615 ÷ 612 = 615-12 = 63

(ଗ) a3 × a7
ସମାଧାନ:
a3 × a7 = a3+7 = a10

(ଘ) 7 × 72
ସମାଧାନ:
7 × 72 = 71 × 72 = 71+2 = 73

(ଙ) 52 ÷ 53
ସମାଧାନ:
52 ÷ 53 = 52-3 = 5-1

(ଚ) 25 × 35
ସମାଧାନ:
25 × 35 = (2 × 3)5 = 65

(ଛ) a4 × a5
ସମାଧାନ:
a4 × a5 = a4+5 = a9

(ଜ) (34)3 × (26)2
ସମାଧାନ:
(34)3 × (26)2 = 34×3 × 26×2 = 312 × 212 = (3 × 2)12 = 612

(ଝ) (210 ÷ 28) × 23
ସମାଧାନ:
(210 ÷ 28) × 23 = 210-8 × 23 = 22 × 23 = 22+3 = 25

Question 2.
ସରଳ କରି ଏକ ଘାତରାଶିରେ ପରିଣତ କର ।

(କ) $$\frac{2^3 \times 3^4 \times 4}{3 \times 3^3}$$
ସମାଧାନ:
$$\frac{2^3 \times 3^4 \times 4}{3 \times 3^3}$$ = $$\frac{2^3 \times 3^4 \times 4}{3^{1+3}}=\frac{2^3 \times 3^4 \times 4}{3^4}$$ = 23 × 4 = 23 × 22 = 23+2 = 25

(ଖ) $$\frac{3 \times 7 \times 11^8}{21 \times 11^3}$$
ସମାଧାନ:
$$\frac{3 \times 7 \times 11^8}{21 \times 11^3}$$ = $$\frac{21 \times 11^8}{21 \times 11^3}=\frac{11^8}{11^3}$$ = 118-3 = 115

(ଗ) [(52)3 × 54] ÷ 57
ସମାଧାନ:
[(52)3 × 54] ÷ 57=[52×3 × 54] + 57
= [56 × 54] + 57 = 56+4 ÷ 57 = 510 ÷ 57 = 510-7 = 53

(ଘ) 254 ÷ 53
ସମାଧାନ:
254 ÷ 53 = (52)4 ÷ 53 = 52×4 + 53 = 58 ÷ 53 = 58-3 = 55

(ଙ) $$\frac{3^7}{\mathbf{3}^4 \times \mathbf{3}^3}$$
ସମାଧାନ:
$$\frac{3^7}{\mathbf{3}^4 \times \mathbf{3}^3}$$ = $$\frac{3^7}{3^{4+3}}=\frac{3^7}{3^7}$$ = 37-7 = 30

(ଚ) $$\frac{2^4 \times a^53}{4^2 \times a}$$
ସମାଧାନ:
$$\frac{2^4 \times a^53}{4^2 \times a}$$ = $$\frac{16 \times a^5}{16 \times a}=\frac{a^5}{a}$$ = a5-1 = a4

(ଛ) (23 × 2)2 ÷ 25
ସମାଧାନ:
(23 × 2)2 ÷ 25 = (23+1)2 ÷ 25 =(24)2 ÷ 25
= 24×2 ÷ 25 = 28 ÷ 25 = 28-5 = 23

(ଜ) $$\frac{a^5}{a^3}$$ × a8
ସମାଧାନ:
$$\frac{a^5}{a^3}$$ × a8 = (a5-3) × a8 = a2 × a8 = a2+8 = a10

Question 3.
ନିମ୍ନ ସଂଖ୍ୟାମାନଙ୍କୁ ମୌଳିକ ସଂଖ୍ୟା ଆଧାର ବିଶିଷ୍ଟ ଏକାଧ୍ଵ ଘାତରାଶିର ଗୁଣଫଳ ରୂପେ ପ୍ରକାଶ କର ।

(କ) 270
ସମାଧାନ:
270 = 2 × 3 × 3 × 3 × 5 = 21 × 33 × 51

(ଖ) 768
ସମାଧାନ:
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 31

(ଗ) 108 × 192
ସମାଧାନ:
108 × 192 = (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= 22 × 33 × 26 × 3 = (22 × 26) × (33 × 3) = 22+6 × 33+1 = 28 × 34

(ଘ) 729 × 64
ସମାଧାନ:
729 × 64 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 = 36 × 26

Question 4.
ସରଳ କର :

(କ) {(42)}2
ସମାଧାନ:
{(4)2}2 = 42×2 = 44 = 4 × 4 × 4 × 4 = 256

(ଖ) (6)3 ÷ (6)
ସମାଧାନ:
(6)3 ÷ (6) = 63-1 = 62 = 36

(ଗ) (2)3 × (3)3 ÷ (6)3
ସମାଧାନ:
(2)3 × (3)3 ÷ (6)3 = $$\frac{2^3 \times 3^3}{6^3}=\frac{(2 \times 3)^3}{6^3}=\frac{6^3}{6^3}$$ = 1

(ଘ) (5)2 × (5)4 ÷ (5)2
ସମାଧାନ:
(5)2 × (5)4 ÷ (5)2 = 52 × 54-2 = 52 × 52 = 52+2 = 54 = 625

(ଙ) $$\frac{\left(2^5\right) \times 7^3}{8^3 \times 7}$$
ସମାଧାନ:
$$\frac{\left(2^5\right) \times 7^3}{8^3 \times 7}$$ = $$\frac{2^5 \times 7^3}{\left(2^3\right)^3 \times 7}=\frac{2^5 \times 7^3}{2^9 \times 7}=\frac{2^5 \times 7^2 \times 7}{2^5 \times 2^4 \times 7}=\frac{7^2}{2^4}=\frac{49}{16}$$

(ଚ) $$\frac{3^2 \times 10^5 \times 25}{5^3 \times 6^4}$$
ସମାଧାନ:
$$\frac{3^2 \times 10^5 \times 25}{5^3 \times 6^4}$$ = $$\frac{3^2 \times(2 \times 5)^5 \times 5^2}{5^3 \times(2 \times 3)^4}=\frac{3^2 \times 2^5 \times 5^5 \times 5^2}{5^3 \times 2^4 \times 3^4}$$

$$=\frac{2^5}{2^4} \times \frac{3^2}{3^2\cdot 3^2} \times \frac{5^{5+2}}{5^3}$$ = $$2^{5-4} \times \frac{1}{3^2} \times 5^{7-3}$$ = $$=2 \times\frac{1}{9} \times 5^4=\frac{2}{9} \times 625$$ = $$\frac{1250}{9}$$