How do you simplify #(1/2) ^ (5/2) #?

1 Answer
Neefgi
Mar 20, 2018

#sqrt(2)/8#

Explanation:

When you have an exponent that is a fraction, in this case, 5/2, the numerator becomes the exponent of the original value (1/2), and the denominator becomes the index of a radical (the expression above the radical symbol)

Obviously those are confusing terms that don't really mean anything; an easy way to remember it is that the numerator moves down, and the denominator moves to the side:

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

In this case, the index is 2. When the index is 2, the expression is just "the square root of," so you don't need to write the index if and only if it's 2, but it's there for clarity.

To simplify it even further, we can evaluate:

#(1/2)^5# is the same as #((1^5)/(2^5))#

Since #1^5 = 1*1*1*1*1 = 1# and
#2^5 = 2*2*2*2*2 = 32#,

#((1^5)/(2^5)) = (1/32)#

Now we are left with #sqrt(1/32)#

To simplify this even further, you would have to simplify #sqrt(32)#:

#sqrt(32) = sqrt(8*4)# Find a perfect square of 32, in this case 4
#2sqrt(8)# We can take the 2 out since the square root of 4 is 2
#2sqrt(4*2)# The square root of 8 can be simplified further
#2*2sqrt(2)# We can take a 2 out again, as we did two steps ago, and multiply it by the 2 that's already out there
#4sqrt(2)#

Don't forget that 1 is a perfect square! Now we have:

#1/(4sqrt(2))#

Unfortunately, we can't leave radicals in the denominator, so we have to multiply the top and bottom by #sqrt(2)#

#1/(4sqrt(2)) * sqrt(2)/sqrt(2)#

#(1sqrt(2))/(4*2)# Don't forget that #sqrt(2)*sqrt(2) = 2#

#sqrt(2)/8# (Don't forget #1sqrt(2)=sqrt(2)#)

Related questions
See all questions in Fractional Exponents
Impact of this question
6964 views around the world
You can reuse this answer
Creative Commons License