Advertisements
Advertisements
Question
Write \[\left( \frac{1}{9} \right)^{- 1/2} \times (64 )^{- 1/3}\] as a rational number.
Advertisements
Solution
We have to find the value of . `(1/9) ^((-1)/2) xx (64) ^((-1)/3`So,
`(1/9) ^((-1)/2) xx (64) ^((-1)/3` = `(1/9) ^((-1)/2) xx (64) ^((-1)/3`,
`= (1/3) ^((-1)/2) xx (4^3) ^((-1)/3) `
`= (1/3^(2 xx (-1)/2)) xx (4^(3 xx (-1)/3))`
`= (1/3^(2 xx (-1)/2)) xx (4^(3 xx (-1)/3))`
`(1/9) ^((-1)/2) xx (64) ^((-1)/3 ) = 1/3^(-1) xx 4^(-1) `
`=1/(1/3) xx 1/4`
`= 1xx 3/1 xx 1/4`
`= 3/4`
Hence the value of the value of `(1/9)^(-1/2) xx (64)^(-1/3)` is `3/4`.
APPEARS IN
RELATED QUESTIONS
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
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`
Prove that:
`(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Solve the following equation:
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
If \[x + \sqrt{15} = 4,\] then \[x + \frac{1}{x}\] =
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
Find:-
`32^(2/5)`
Find:-
`125^((-1)/3)`
