Number System: 9 Math
NCERT Exercise 1.6 Laws of Exponents Real Numbers: 9th math
Laws of Exponents for Real Numbers
(i) `a^m\ *\ a^n=a^(m+n)`
(ii) `(a^m)^n = a^(m*n)`
(iii) `a^m/a^n = a^(m-n)`, m>n
(iv) `a^m\ b^m = (a\ b)^m`
(v) `a^0 = 1`
(vi) `1/a^n=a^(-n)`
Solution of NCERT Exercise 1.6 (class nine mathematics CBSE
course)Question (1) Find
(i) `64^(1/2)`
Solution
Given, `64^(1/2)`
`=(8xx8)^(1/2)`
`=(8^2)^(1/2)`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
∴ `(8^2)^(1/2) =8^(2xx1/2)`
`=8^1 = 8`
Thus, `64^(1/2)= 8` Answer
(ii) `32^(1/5)`
Solution:
Given, `32^(1/5)`
`=(2xx2xx2xx2xx2)^(1/5)`
`=(2^5)^(1/5)`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
∴ `=(2^5)^(1/5) = 2^(5xx1/5)`
`=2^1 = 2`
Thus, `32^(1/5)=2` Answer
(iii) `125^(1/3)`
Solution:
Given, `125^(1/3)`
`=(5xx5xx5)^(1/3)`
`=(5^3)^(1/3)`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
∴ `(5^3)^(1/3)=5^(3xx1/3)`
`=5^1=5`
Thus, `125^(1/3) =5` Answer
Question (2) Find
(i) `9^(3/2)`
Solution:
Given, `9^(3/2)`
`=(3xx3)^(3/2)`
`=(3^2)^(3/2)`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
∴ `(3^2)^(3/2)=3^(2xx3/2)`
`=3^3=3xx3xx3=27`
Thus, `9^(3/2)=27` Answer
(ii) `32^(2/5)`
Solution:
Given, `32^(2/5)`
`=(2xx2xx2xx2xx2)^(2/5)`
`=(2^5)^(2/5)`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
∴ `(2^5)^(2/5)=2^(5xx2/5)`
`=2^2=2xx2=4`
Thus, `32^(2/5) =4` Answer
(iii) `16^(3/4)`
Solution:
Given, `16^(3/4)`
`=(2xx2xx2xx2)^(3/4)`
`=(2^4)^(3/4)`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
∴ `(2^4)^(3/4) = 2^(4xx3/4)`
`=2^3=2xx2xx2=8`
Thus, `16^(3/4) = 8` Answer
(iv) `125^((-1)/3)`
Solution:
Given, `125^((-1)/3)`
`= (5xx5xx5)^ ((-1)/3)`
`=(5^3)^((-1)/3)`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
Therefore, `(5^3)^((-1)/3)=5^(3xx(-1)/3)`
`=5^(-1)`
Now, according to Laws of Exponents we know that, `1/a^m= a^(-m)`
Thus, `5^(-1)=1/5`
Therefore, `125^((-1)/3)=1/5` Answer
Question (3) Simplify:
(i) `2^(2/3)\ *\ 2^(1/5)`
Solution:
Given, `2^(2/3)\ *\ 2^(1/5)`
According to Laws of Exponents, we know that, `a^m\ * \ a^n=a^(m+n)`
Therefore, `2^(2/3)\ *\ 2^(1/5)`
`=2^(2/3+1/5)`
`=2^((10+3)/15)`
`=2^(13/15)` Answer
(ii) `(1/3^3)^7`
Solution :
Given, `(1/3^3)^7`
According to Laws of Exponents, we know that, `1/a^n = a^(-n)`
`=(3^(-3))^7`
According to Laws of Exponents, we know that, `(a^m)^n=a^(m*n)`
Therefore, `(3^(-3))^7 = 3^(-3xx7)`
`=3^(-21)` Answer
Or, `=1/3^(-21)` Answer
(iii) `11^(1/2)/11^(1/4)`
Solution:
Given, `11^(1/2)/11^(1/4)`
According to Laws of Exponents, we know that `a^m/a^n = a^(m-n)`. Thus by applying this law to the given expression we get
`11^(1/2-1/4)`
`=11^((2-1)/4)`
`=11^(1/4)` Answer
(iv) `7^(1/2)\ *\ 8^(1/2)`
Solution:
Given, `7^(1/2)\ *\ 8^(1/2)`
According to Laws of Exponents, we know that, `a^m\ *\ b^m=(ab)^m`. Thus by applying this law to the given expression, we get
`(7xx8)^(1/2)`
`=56^(1/2)` Answer
Reference: