Number System

Exercise 1.6

Important Laws on Indices:

Question 1: Find

(i) `64^(1/2)`

Answer: Given, `64^(1/2)`

`=(8xx8)^(1/2)=(8^2)^(1/2)`

`=8^(2xx1/2)=8^1=8`


(ii) `32^(1/5)`

Answer: Given, `32^(1/5)`

`=(2xx2xx2xx2xx2)^(1/5)`

`=2^(5xx1/5)=2^1=2`

(iii) `125^(1/3)`

Answer: Given, `125^(1/3)`

`=(5xx5xx5)^(1/3)`

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


Question 2: Find

(i) `9^(3/2)`

Answer: Given, `9^(3/2)`

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

`=3^(2xx3/2)=3^3=27`

(ii) `32^(2/5)`

Answer: Given, `32^(2/5)`

`=(2xx2xx2xx2xx2)^(2/5)`

`=(2^5)^(2/5)=2^(5xx2/5)=2^2=4`

(iii) `16^(3/4)`

Answer: Given, `16^(3/4)`

`=(2xx2xx2xx2)^(3/4)`

`=(2^4)^(3/4)=2^(4xx3/4)=2^3=8`

(iv) `125^(-1/3)`

Answer: Given, `125^(-1/3)`

`=(5xx5xx5)^(-1/3)`

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

Question 3: Simplify

(i) `2^(2/3).2^(1/5)`

Answer: Given, `2^(2/3).2^(1/5)`

Since, `a^m.a^n=a^(m+n)`

`=(2)^(2/3+1/5)=(2)^((10+3)/15)=2^(13/15)`

(ii) `(1/(3^3))^7`

Answer: Given, `(1/(3^3))^7`

`=(3^-3)^7=3^(-3xx7)=3^-21`

(iii) `(11^(1/2))/(11^(1/4))`

Answer: Given, `(11^(1/2))/(11^(1/4))`

`=(11)^(1/2-1/4)=(11)^((2-1)/4)=11^1/4`

(iv) `7^(1/2).8^(1/2)`

Answer: Given, `7^(1/2).8^(1/2)`

We know that, `a^m.b^m=(ab)^m`

`=(7xx8)^(1/2)=56^(1/2)`



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