Advertisements
Advertisements
Question
If x is a positive real number and x2 = 2, then x3 =
Options
\[\sqrt{2}\]
2\[\sqrt{2}\]
3\[\sqrt{2}\]
4
Advertisements
Solution
We have to find `x^3`provided `x^2 = 2`. So,
By raising both sides to the power `1/2`
`x^(2 xx 1/2) = 2^(1/2)`
`x^(2 xx 1/2) = sqrt2`
`x= sqrt2`
By substituting `x= sqrt2` in `x^2` we get
`x^2 = (sqrt2)^3`
= `sqrt2 xx sqrt2 xxsqrt2`
= `2sqrt2`
The value of `x^2`is `2sqrt2`
APPEARS IN
RELATED QUESTIONS
Simplify the following:
`(2x^-2y^3)^3`
Solve the following equation for x:
`7^(2x+3)=1`
Simplify:
`root3((343)^-2)`
Prove that:
`9^(3/2)-3xx5^0-(1/81)^(-1/2)=15`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
If a and b are different positive primes such that
`(a+b)^-1(a^-1+b^-1)=a^xb^y,` find x + y + 2.
Which one of the following is not equal to \[\left( \sqrt[3]{8} \right)^{- 1/2} ?\]
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
