How do you simplify #(y^6/y) ^-2# using only positive exponents?

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Answer 1 · 2016-08-18T16:56:27

#1/y^10#

#((cancel(y)xxy^5)/(cancel(y)))^(-2)#

#1/((y^5)^2)#

#1/y^10#

Answer 2 · 2016-08-18T16:56:35

#=1/y^10#

#(y^6/y)^-2#

#=(y^5)^-2#

#=y^-10#

#=1/y^10#

Answer 3 · 2016-08-18T17:00:11

#1/y^10#

first observe that #y^(x^n) = y^(xn)# and #(y^n/y^1)=y^(n-1)#so

#(y^6/y)^-2=y^-10#

when an exponent is negative this is the same as placing the positive version under 1

#y^-10 = 1/y^10#.

Answer 4 · 2016-08-24T06:36:39

#1/y^10#

There is a useful variation of one of the laws of indices.

Note: #color(blue)((a/b)^-c = (b/a)^(+c)) #

#(y^6/y)^color(red)(-2) = (y/y^6)^color(red)(2)#

There are no longer negative indices, simplify as normal.

=# (1/y^5)^2#

= #1/y^10#