Advertisements
Advertisements
Question
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
Options
5a2bc2
25ab2c
5a3bc3
125a2bc2
Advertisements
Solution
Find value of \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\]
\[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] = `5sqrt(5^5 a^10 b^5 c^10)`
`= 5^(5 xx 1/5) a^(10 xx 1/5 ) b^(5 xx 1/5 ) c^(10xx1/5)`
`= 5^(5 xx 1/5) a^(10 xx 1/5 ) b^(5 xx 1/5 ) c^(10xx1/5)`
\[\sqrt[5]{3125 a^{10} b^5 c^{10}} = 5 a^2 b c^2\]
APPEARS IN
RELATED QUESTIONS
Simplify the following
`(4ab^2(-5ab^3))/(10a^2b^2)`
Simplify:
`(16^(-1/5))^(5/2)`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
If `5^(3x)=125` and `10^y=0.001,` find x and y.
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
If 9x+2 = 240 + 9x, then x =
If \[4x - 4 x^{- 1} = 24,\] then (2x)x equals
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
