Advertisements
Advertisements
प्रश्न
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
पर्याय
5a2bc2
25ab2c
5a3bc3
125a2bc2
Advertisements
उत्तर
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
संबंधित प्रश्न
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
Solve the following equation for x:
`2^(3x-7)=256`
Prove that:
`(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
Find the value of x in the following:
`2^(5x)div2x=root5(2^20)`
Simplify:
`(x^(a+b)/x^c)^(a-b)(x^(b+c)/x^a)^(b-c)(x^(c+a)/x^b)^(c-a)`
If `a=x^(m+n)y^l, b=x^(n+l)y^m` and `c=x^(l+m)y^n,` Prove that `a^(m-n)b^(n-l)c^(l-m)=1`
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
If 9x+2 = 240 + 9x, then x =
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
Simplify:
`11^(1/2)/11^(1/4)`
