Advertisements
Advertisements
Question
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
Options
\[\left\{ \left( \frac{5}{6} \right)^\frac{1}{5} \right\}^{- \frac{3}{6}}\]
\[\frac{1}{\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{1/6}}\]
\[\left( \frac{6}{5} \right)^{1/30}\]
\[\left( \frac{5}{6} \right)^{- 1/30}\]
Solution
We have to find the value of `{(5/6)^(1/5)}^((-1) / 6)`
So,
`{(5/6)^(1/5)}^((-1) / 6) = 5^(1/5 xx (-1)/6) / 6^(1/5 xx (-1)/6)`
`=5^(-1/30)/(6^((-1)/30))`
`=(1/(5^(-1/30))) / (1/(6^(1/30))`
`{(5/6)^(1/5)}^((-1) / 6)` = `1/(5^(1/30)) xx (6^(1/30))/1`
= `(6^(1/30))/5^(1/30)`
= `(6/5)^(1/30)`
APPEARS IN
RELATED QUESTIONS
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
Determine `(8x)^x,`If `9^(x+2)=240+9^x`
Solve the following equation:
`3^(x+1)=27xx3^4`
If a and b are distinct primes such that `root3 (a^6b^-4)=a^xb^(2y),` find x and y.
Simplify:
`root(lm)(x^l/x^m)xxroot(mn)(x^m/x^n)xxroot(nl)(x^n/x^l)`
If x-2 = 64, then x1/3+x0 =
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
Simplify:
`7^(1/2) . 8^(1/2)`