Advertisements
Advertisements
प्रश्न
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
Advertisements
उत्तर
`((x^2y^2)/(a^2b^3))^n`
`=((x^2)^n(y^2)^n)/((a^2)^n(b^3)^n)`
`=(x^(2n)y^(2n))/(a^(2n)b^(3n))`
APPEARS IN
संबंधित प्रश्न
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Find the value of x in the following:
`(3/5)^x(5/3)^(2x)=125/27`
Simplify:
`(x^(a+b)/x^c)^(a-b)(x^(b+c)/x^a)^(b-c)(x^(c+a)/x^b)^(c-a)`
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
(256)0.16 × (256)0.09
If \[4x - 4 x^{- 1} = 24,\] then (2x)x equals
If \[\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x},\] then x =
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
