Number System
Exercise 1.6
Important Laws on Indices:
- `a^m.a^n=a^(m+n)`
- `(a^m)^n=a^(mn)`
- `(a^m)/(a^n)=a^(m-n)`
- `a^m\b^m=(ab)^m`
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)`