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) $\frac{64}{729}=\frac{2^6}{3^6}=\left(\frac{2}{3}\right)^6=\left(\frac{3}{2}\right)^{-6}$