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
If a = 3 and b = -2, find the values of :
ab + ba
Prove that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
If `27^x=9/3^x,` find x.
Find the value of x in the following:
`5^(2x+3)=1`
The square root of 64 divided by the cube root of 64 is
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 \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
