kaylala
(3^2n times a^3)^2 = ?
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TjRoDz
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hmm does it have choices?
my answer is 9a^6n
but i could be wrong
kaylala
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\[(3^{4n}a ^{3})^{2^{}} = ???\]
TjRoDz
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ok sorry i was wrong
kaylala
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@TjRoDz
you're wrong...
TjRoDz
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uhh 81a^6n
kaylala
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still wrong @TjRoDz
TjRoDz
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hhhaha my pea sized brain
kaylala
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help me please
@thomaster
thomaster
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uhm i think \(\sf\Large a^6\large3^{\Large8n}\)
thomaster
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\(\Large (a^n)^m=a^{n*m}\)
kaylala
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wont it be instead of 3 it would be squared and be 9?
kaylala
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@thomaster
wont it be:
instead of 3; it would be squared and be 9?
kaylala
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i'm confused
thomaster
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No
\(\Large(3^{4n}a ^3)^2\) 4*2=8 and 3*2=6
\(\Large3^{8n}a ^6~\to~a^63^{8n}\)
phi
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you can memorize the rule
\[ (x^a)^b = x^{ab} \]
which means multiply the exponents
but you can use this idea:
\[ x^2 \text{ means } x \cdot x \]
so
\[ (3^{4n}a ^{3})^{2} \text{ means } (3^{4n}a ^{3})(3^{4n}a ^{3}) \]
phi
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you can re-arrange that to
\[ (3^{4n}\cdot 3^{4n}\cdot a ^{3}\cdot a ^{3})\]
phi
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if you don't know about "adding exponents" you could figure out
\[ a^3 \cdot a^3 \]
by knowing that \(a^3 = a\cdot a \cdot a \)
\[ a^3 \cdot a^3 = a\cdot a \cdot a \cdot a\cdot a \cdot a= a^6\]
phi
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but to do
\[ 3^{4n} \cdot 3^{4n} \]
you need to know the rule: if you have the same base, add the exponents
\[ 3^{4n+4n} \]
or
\[3^{8n} \]