# BSE Odisha 8th Class Maths Solutions Algebra Chapter 5 ସୂଚକ ତତ୍ତ୍ଵ Ex 5(c)

Odisha State Board BSE Odisha 8th Class Maths Solutions Algebra Chapter 5 ସୂଚକ ତତ୍ତ୍ଵ Ex 5(c) Textbook Exercise Questions and Answers.

## BSE Odisha Class 8 Maths Solutions Algebra Chapter 5 ସୂଚକ ତତ୍ତ୍ଵ Ex 5(c)

Question 1.
ନିମ୍ନଲିଖ ରାଶିଗୁଡ଼ିକୁ ଏକ ଆଧାର ବିଶିଷ୍ଟ ଘାତ ରାଶି ରୂପେ ପ୍ରକାଶ କର ।
(i) 2-2
(ii) 2-4
(iii) 3-3
(iv) 3-5
(v) 10-4
(vi) 5-3
(vii) 20-3
(viii) 50-3
(ix) 100-1
(x) (0.1)5
(xi) (-1)-1
(xii) (-1)-27
ସମାଧାନ :
(i) 2-2 = $$\frac{1}{2^2}=\frac{1}{4}$$
(ii) 2-4 = $$\frac{1}{2^4}=\frac{1}{16}$$
(iii) 3-3 = $$\frac{1}{3^3}=\frac{1}{27}$$
(iv) 3-5 = $$\frac{1}{3^5}=\frac{1}{243}$$
(v) 10-4 = $$\frac{1}{10^4}=\frac{1}{10000}$$
(vi) 5-3 = $$\frac{1}{5^3}=\frac{1}{125}$$
(vii) 20-3 = $$\frac{1}{20^3}=\frac{1}{8000}$$
(viii) 50-3 = $$\frac{1}{50^3}=\frac{1}{125000}$$
(ix) 100-1 = $$\frac{1}{100^1}=\frac{1}{100}$$
(x) (0.1)5 = $$\left(\frac{1}{10}\right)^5=\frac{1}{10^5}=\frac{1}{100000}$$
(xi) (-1)-1 = $$\frac{1}{(-1)^1}=\frac{1}{-1}=-1$$
(xii) (-1)-27 = $$\frac{1}{(-1)^{27}}=\frac{1}{-1}=-1$$

Question 2.
ସରଲ କର ।
(i) $$\left(\frac{1}{3}\right)^{-2}$$
(ii) $$\left(\frac{2}{5}\right)^{-3}$$
(iii) $$\left(\frac{1}{10}\right)^{-4}$$
(iv) (0.2)3
(v) $$\left(\frac{3}{5}\right)^{-3}$$
(vi) $$\left(\frac{3}{10}\right)^{-3}$$
(vii) (-1)-101
(viii) (-1)1000
ସମାଧାନ :
(i) $$\left(\frac{1}{3}\right)^{-2}=\frac{1}{\left(\frac{1}{3}\right)^2}=\frac{1}{\frac{1}{9}}=9$$
(ii) $$\left(\frac{2}{5}\right)^{-3}=\frac{1}{\left(\frac{2}{5}\right)^3}=\frac{1}{\frac{8}{125}}=\frac{125}{8}$$
(iii) $$\left(\frac{1}{10}\right)^{-4}=\frac{1}{\left(\frac{1}{10}\right)^4}=\frac{1}{\frac{1}{10000}}=10000$$
(iv) (0.2)3 = $$\left(\frac{2}{10}\right)^3=\left(\frac{1}{5}\right)^3=\frac{1}{125}$$
(v) $$\left(\frac{3}{5}\right)^{-3}=\left(\frac{5}{3}\right)^3=\frac{125}{27}$$
(vi) $$\left(\frac{3}{10}\right)^{-3}=\left(\frac{10}{3}\right)^3=\frac{1000}{27}$$
(vii) $$(-1)^{-101}=\frac{1}{(-1)^{101}}=\frac{1}{-1}=-1$$
(viii) (-1)1000 = 1

Question 3.
ମୌଳିକ ଆଧାର ବିଶିଷ୍ଟ ଘାତରାଶିରେ ପ୍ରକାଶ କର ।
(i) 36
(ii) (6)3
(iii) -216
(iv) 625
(v) 343
(vi) $$\frac{1}{512}$$
(vii) $$\frac{64}{729}$$
ସମାଧାନ :
(i) 36 = $$\frac{1}{3^{-6}}=\left(\frac{1}{3}\right)^{-6}$$
(ii) (6)3 = $$\frac{1}{6^{-3}}=\left(\frac{1}{6}\right)^{-3}$$
(iii) -216 = $$(-6)^3=\left(\frac{-6}{1}\right)^3=\left(\frac{1}{-6}\right)^{-3}$$
(iv) 625 = $$5^4=\frac{1}{5^{-4}}=\left(\frac{1}{5}\right)^{-4}$$
(v) 343 = $$7^3=\frac{1}{7^{-3}}=\left(\frac{1}{7}\right)^{-3}$$
(vi) $$\frac{1}{512}=\frac{1}{2^9}=2^{-9}$$
(vii) $$\frac{64}{729}=\frac{2^6}{3^6}=\left(\frac{2}{3}\right)^6=\left(\frac{3}{2}\right)^{-6}$$