Algebra 1: Common Core (15th Edition)

The answer is $a^{-4}$
To solve this $a^{0}$=1 as any number with exponent 0 becomes 1 $a^{x}$ $\times$ $a^{4}$ = 1 $a^{x}$ $\times$ $a^{4}$ = $a^{0}$ $a^{x}$ = $\frac{a^{0}}{a^{4}}$ $a^{x}$ = $a^{0-4}$ $a^{x}$ = $a^{-4}$ --> same base, bring the exponents down x= -4 Therefore the missing part is $a^{-4}$.