How do you evaluate #[(7-5)^5/8]-4#?

1 Answer
Jun 16, 2015

The answer is zero.

Explanation:

First, note that the parentheses mean that #7-5# should be computed first to get #7-5=2#. Then raise that to the fifth power to get #(7-5)^5=2^5=32# (powers come before divisions in order of operations). Then divide by 8 to get #((7-5)^2)/8=32/8=4#. Finally, subtract 4 to get #((7-5)^2)/8-4=4-4=0#.

The outer brackets #[ ]# are not technically needed in the notation.