How do you evaluate #32# to the power of #2/5#? #32^(2/5)#

1 Answer
EZ as pi
Aug 25, 2016

#4#
There is a choice of methods.. Work smarter, not harder!

Explanation:

One of the laws of indices deals with cases where there are powers and roots at the same time.

#x^(p/q) = rootq(x^p) = (rootq x)^p #

The denominator shows the root and the numerator gives the power.

Note that the power can be inside or outside the root.

I prefer to find the root first, and then raise to the power because this keeps the numbers smaller. They can usually be calculated mentally rather than needing a calculator

#32^(2/5) = (color(red)root5(32))^2#

=#color(red)(2)^2color(white)(wwwwwwwwwwwww)(2*2*2*2*2=2^5=32)#

=#4#

Compare this with the other method of squaring first.

#root5(color(blue)(32^2)) = root5(color(blue)(1024))#

=#4#

While I know that #2^5 = 32 #, the square of #32# and the fifth root of #1024# are not facts that I would be able to recall from memory.

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