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joyce153
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Hunus
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\[a^{12}*a^{0}=a^{12}*1=a^{12}\]

meverett04
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What can we do with the 12 and 0?
Hint: \[x^3 \times x^5 = x^8\]

joyce153
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so the answer is \[a ^{12}\]

Hunus
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Yes, because
\[a^{m}*a^{n}=a^{m+n}\]\[a^{12}*a^{0}=a^{12+0}=a^{12}\]

joyce153
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kk THX

Hunus
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no problem :)