How do you evaluate #-4^{-3+}2^{0}#?

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Answer 1 · 2017-11-28T02:47:04

See a solution process below:

First, use this rule of exponents to rewrite the 2 term:

#a^color(red)(0) = 1#

#-4^-3 + 2^color(red)(0) => -4^-3 + 1#

Next, use this rule of exponents to rewrite the #4# term:

#x^color(red)(a) = 1/x^color(red)(-a)#

#-4^color(red)(-3) + 1 => -1/4^color(red)(- -3) + 1 =>#

#-1/4^color(red)(3) + 1 => -1/64 + 1#

Now, put each term over a common denominator to add the terms:

# -1/64 + (64/64 xx 1) => (-1)/64 + 64/64 => (-1 + 64)/64 = 63/64#