Advertisements
Advertisements
प्रश्न
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
विकल्प
5a2bc2
25ab2c
5a3bc3
125a2bc2
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
`(2x^-2y^3)^3`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Find the value of x in the following:
`5^(x-2)xx3^(2x-3)=135`
If `5^(3x)=125` and `10^y=0.001,` find x and y.
Solve the following equation:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
Solve the following equation:
`sqrt(a/b)=(b/a)^(1-2x),` where a and b are distinct primes.
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`
If 102y = 25, then 10-y equals
If \[2^{- m} \times \frac{1}{2^m} = \frac{1}{4},\] then \[\frac{1}{14}\left\{ ( 4^m )^{1/2} + \left( \frac{1}{5^m} \right)^{- 1} \right\}\] is equal to
If o <y <x, which statement must be true?