Advertisements
Advertisements
प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`root5(243x^10y^5z^10)`
उत्तर
We have to simplify the following, assuming that x, y, z are positive real numbers
Given `root5(243x^10y^5z^10)`
`=(243xx x^10xxy^5xxz^10)^(1/5)`
`=(243)^(1/5)xx (x^10)^(1/5)xx(y^5)^(1/5)xx(z^10)^(1/5)`
`=(3^5)^(1/5)xx x^(10xx1/5)xxy^(5xx1/5)xxz^(10xx1/5)`
`=3xx x^2xxyxxz^2`
`=3x^2yz^2`
APPEARS IN
संबंधित प्रश्न
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
Determine `(8x)^x,`If `9^(x+2)=240+9^x`
If `3^(4x) = (81)^-1` and `10^(1/y)=0.0001,` find the value of ` 2^(-x+4y)`.
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)`
`(2/3)^x (3/2)^(2x)=81/16 `then x =
The value of \[\left\{ \left( 23 + 2^2 \right)^{2/3} + (140 - 19 )^{1/2} \right\}^2 ,\] is
If \[x + \sqrt{15} = 4,\] then \[x + \frac{1}{x}\] =
The positive square root of \[7 + \sqrt{48}\] is
Find:-
`125^(1/3)`