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kaylala

(3^2n times a^3)^2 = ?

  • 9 months ago
  • 9 months ago

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  1. TjRoDz
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    hmm does it have choices? my answer is 9a^6n but i could be wrong

    • 9 months ago
  2. kaylala
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    \[(3^{4n}a ^{3})^{2^{}} = ???\]

    • 9 months ago
  3. TjRoDz
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    ok sorry i was wrong

    • 9 months ago
  4. kaylala
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    @TjRoDz you're wrong...

    • 9 months ago
  5. TjRoDz
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    uhh 81a^6n

    • 9 months ago
  6. kaylala
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    still wrong @TjRoDz

    • 9 months ago
  7. TjRoDz
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    hhhaha my pea sized brain

    • 9 months ago
  8. kaylala
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    help me please @thomaster

    • 9 months ago
  9. thomaster
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    uhm i think \(\sf\Large a^6\large3^{\Large8n}\)

    • 9 months ago
  10. thomaster
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    \(\Large (a^n)^m=a^{n*m}\)

    • 9 months ago
  11. kaylala
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    wont it be instead of 3 it would be squared and be 9?

    • 9 months ago
  12. kaylala
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    @thomaster wont it be: instead of 3; it would be squared and be 9?

    • 9 months ago
  13. kaylala
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    i'm confused

    • 9 months ago
  14. thomaster
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    No \(\Large(3^{4n}a ^3)^2\) 4*2=8 and 3*2=6 \(\Large3^{8n}a ^6~\to~a^63^{8n}\)

    • 9 months ago
  15. 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 \[ (3^{4n}a ^{3})^{2} \text{ means } (3^{4n}a ^{3})(3^{4n}a ^{3}) \]

    • 9 months ago
  16. phi
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    you can re-arrange that to \[ (3^{4n}\cdot 3^{4n}\cdot a ^{3}\cdot a ^{3})\]

    • 9 months ago
  17. 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 \(a^3 = a\cdot a \cdot a \) \[ a^3 \cdot a^3 = a\cdot a \cdot a \cdot a\cdot a \cdot a= a^6\]

    • 9 months ago
  18. 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 \[3^{8n} \]

    • 9 months ago
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