What is #((8m^5n^7)/(2mn^5))^3#?

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Answer 1 · 2018-03-26T09:41:38

#64m^12n^6#

Just Use Laws of Indices and you're done.

We have,

#color(white)(xx)((cancel8^4 xxm^5n^7)/(cancel2xxmn^5))^3#

#= ((4m^5n^7)/(mn^5))^3#

#= (4m^5n^7)^3/(mn^5)^3# [As #(a/b)^m = a^m/b^m#]

#= (64m^15n^21)/(m^3n^15)# [As #(a^m)^n = a^(mn)#]

#= 64m^15n^21 xx m^-3n^-15# [As #a^-m = 1/a^m#]

#= 64 xx m^(15 +(-3)) n^(21 +(-15))# [As #a^(m+n) = a^m xx a^n#]

#= 64 xx m^12 n^6 = 64m^12n^6#

Hence Explained.

Answer 2 · 2018-03-26T09:55:03

#color(blue)(=> (64 m^12 n^6)#

#"Given : " ((8m^5n^7)/(2 m n^5))^3#

![ quantbasics.blogspot.in )

#=> ((2^3 m^5 n^7)/(2 m n^5))#

#=> (2^(3-1) * m^(5-1) * n^(7-5))^3,# # "color(red)(as ( a^m / a^n = a^(m-n))#

#=> (2^2 * m^4 * n^2)^3#

#color(blue)(=> (2^6 m^12 n^6)#, # color(red)("as "( (a^m)^n = a^(mn))#

#color(blue)(=> 64m^(12)n^6#, # " as (" 2^6 = 64)#