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Evaluate : ( 27/125 )^(2/3) Xx ( 9/25 )^(3/2) - Mathematics

Sum

Evaluate :
`( 27/125 )^(2/3) xx ( 9/25 )^(3/2)`

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Solution

`( 27/125 )^(2/3) xx ( 9/25 )^(3/2)`

= `([ 3 xx 3 xx 3 ]/[ 5xx 5 xx 5 ])^(2/3) xx ([ 3 xx 3 ]/[ 5 xx 5 ])^( -3/2 )`

= `[(3/5)^3 ]^(2/3) xx [(3/5)^2]^(-3/2)`

= `(3/5)^( 3 xx 2/3 ) xx (3/5)^( 2 xx - 3/2)`

= `(3/5)^2 xx (3/5)^(-3)`

= `(3/5)^( 2 - 3 )`

= `(3/5)^( -1 )`

= `1/(3/5) = 5/3` 

Concept: Laws of Exponents
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APPEARS IN

Selina Concise Mathematics Class 9 ICSE
Chapter 7 Indices (Exponents)
Exercise 7 (A) | Q 1.3 | Page 98
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