Advertisements
Advertisements
Question
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt(x^-3))^5`
Advertisements
Solution
We have to simplify the following, assuming that x, y, z are positive real numbers
Given `(sqrt(x^-3))^5`
As x is positive real number then we have
`(sqrt(x^-3))^5=(sqrt(1/x^3))^5`
`=(sqrt1/sqrt(x^3))^5`
`=(1/x^(3xx1/2))^5`
`=(1/x^(3/2))`
`(sqrt(x^-3))^5=(1^5/x^(3/2xx5))`
`=1/x^(15/2)`
Hence the simplified value of `(sqrt(x^-3))^5` is `1/x^(15/2)`
APPEARS IN
RELATED QUESTIONS
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
Find the value of x in the following:
`2^(x-7)xx5^(x-4)=1250`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
If (x − 1)3 = 8, What is the value of (x + 1)2 ?
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
The positive square root of \[7 + \sqrt{48}\] is
