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2.Simplify and express the result in power notation with a positive exponent. (i)\((-4)^5÷(-4)^8\)
(ii)\((\frac{1}{2^3})^2\)

(iii)\((-3)^4\times (\frac{5}3)^4\)

(iv)\((3^{-7}÷3^{-10})\times 3^{-5}\)

(v)\(2^{-3}\times(-7)^{-3}\)


Answer :

(i)We have:\((-4)^5÷(-4)^8\)
\(=(-4)^{5-8}=(-4)^{-3}=\frac{1}{(-4)^3}\quad\)[∵\(a^{m}÷a^{n}=a^{m-n}\)]

\(=(-\frac{1}4)^3\)

(ii)We have:\((\frac{1}{2^3})^2\)

\((\frac{1}{2^3})^2=\frac{(1)^2}{(2^3)^2}=\frac{1}{2^6}=(\frac{1}2)^6\)

(iii)We have:\((-3)^4\times (\frac{5}3)^4\)

\(=(-3)^4\times (\frac{5}3)^4=(-3)^4\times \frac{(5)^4}{(3)^4}\)

\(=\frac{(3)^4\times(5)^4}{(3)^4}=(5)^4\)

(iv)We have:\((3^{-7}÷3^{-10})\times 3^{-5}\)

\(=(3^{-7}÷3^{-10})\times 3^{-5}=3^{-7-(-10)}\times 3^{-5}\)

\(=3^{-7+10}\times 3^{-5}=3^3\times 3^{-5}=3^{(3-5)}\)

\(=3^{-2}=\frac{1}{3^2}=(\frac{1}3)^2\)

(v)We have:\(2^{-3}\times(-7)^{-3}\)

\(=2^{-3}\times(-7)^{-3}=[2\times(-7)]^{-3}=(-14)^{-3}\)

\(=-(14)^{-3}=-\frac{1}{14^3}=(-\frac{1}{14})^3\)