Advertisements
Advertisements
Question
The product of the square root of x with the cube root of x is
Options
cube root of the square root of x
sixth root of the fifth power of x
fifth root of the sixth power of x
sixth root of x
Solution
We have to find the product (say L) of the square root of x with the cube root of x is. So,
`L = 2sqrtx xx 3sqrtx`
`= x^(1/2) xx x ^(1/3)`
`= x^(1/2+1/3)`
`= x^((1xx3)/(2xx3)+(1xx2)/(3xx2))`
\[= x^\frac{3 + 2}{6} = x^\frac{5}{6}\]
The product of the square root of x with the cube root of x is `x^(5/6)`
Hence the correct alternative is b.
APPEARS IN
RELATED QUESTIONS
Find:-
`9^(3/2)`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Prove that:
`(2^(1/2)xx3^(1/3)xx4^(1/4))/(10^(-1/5)xx5^(3/5))div(3^(4/3)xx5^(-7/5))/(4^(-3/5)xx6)=10`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
If x is a positive real number and x2 = 2, then x3 =
If g = `t^(2/3) + 4t^(-1/2)`, what is the value of g when t = 64?
If \[2^{- m} \times \frac{1}{2^m} = \frac{1}{4},\] then \[\frac{1}{14}\left\{ ( 4^m )^{1/2} + \left( \frac{1}{5^m} \right)^{- 1} \right\}\] is equal to
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
Which of the following is equal to x?
Simplify:
`(1^3 + 2^3 + 3^3)^(1/2)`