How do you find the value of #125^(-2/3)#?

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Answer 1 · 2015-09-23T11:37:25

#125^(-2/3) = 1/25#

#125^(-2/3)#

#=1/(125^(2/3))#

#=1/((125^(1/3))^2)#

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

#=1/25#

Answer 2 · 2015-09-23T11:39:01

#1/25#

According to the law of indices, #a^-m=1/a^m#. So you only need to get the reciprocal to get rid of the negative exponent.

#125^(-2/3)#
#=1/(125^(2/3)#

According to the law of indices again, #a^(m/n)=root(n)(a^m)#.

#1/(125^(2/3)#
#=1/root(3)(125^2)#

Let's factor out 125 completely so we can take out any perfect cubes.

#1/root(3)(125^2)#
#=1/root(3)((5^3)^2)#
#=1/root(3)(5^3*5^3)#
#=1/(5*5)#
#=color(blue)(1/25)#

Answer 3 · 2016-09-26T19:59:12

#1/25#

Note that #125=5^3#.

Thus:

#(125)^(-2/3)=(5^3)^(-2/3)#

Now we should use the rule #(a^b)^c=a^(bxxc)#, so:

#(5^3)^(-2/3)=5^((3xx-2/3))=5^-2#

Here, use the rule #a^-b=1/a^b#:

#5^-2=1/5^2=1/25#