itsjustme_lol
Laws of Exponents..im not understanding this
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LogicalApple
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Like \[a ^{m + n} = a ^{m}a ^{n} ?\]
itsjustme_lol
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idk. hang on let me seee lol
itsjustme_lol
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(3a^4)3
itsjustme_lol
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thats one of them problems it gives me.
KingGeorge
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For that problem, there's two laws that you need to be familiar with.
Law 1:\[\large (a\cdot b)^n=a^n\cdot b^n\]
Law 2: \[\large (a^n)^m=a^{n\cdot m}\]Using these, can you find the solution to your problem.
itsjustme_lol
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ok that makes a little bit more sense . so could you help me solve that problem that I posted?
KingGeorge
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Sure. First, the correct formatting is \[\large (3a^4)^3\]correct?
itsjustme_lol
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yes
itsjustme_lol
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and thankyou!
KingGeorge
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First, we use the first law I mentioned. So \[\large (3a^4)^3=3^3\cdot(a^4)^3\]Now, use law 2 to simplify \((a^4)^3\). Can you tell me what you get?
itsjustme_lol
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ok give me a sec
itsjustme_lol
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(a)^7 ?? i think thats totaly wrong
itsjustme_lol
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unless is a^12
KingGeorge
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It's the second one. Good job!
That means, we've simplified to \[\large 3^3\cdot a^{12} \small .\]Now just find \(3^3\), and you're done.
itsjustme_lol
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with the 3^3 do I multiply that..is it 9?
KingGeorge
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\(3^3=(3\cdot3)\cdot3=9\cdot3=27\)
itsjustme_lol
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oh i see what you did there, i seee my mistake
itsjustme_lol
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so its (a^12)27?
precal
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|dw:1356824001142:dw|also, parenthesis between the exponents remind me to multiply the powers, this is one of the ways I remember this law
KingGeorge
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You could write it as \(27\cdot a^{12}\) or \(a^{12}\cdot 27\).
And if you have some trouble remembering the law, precal's way to do it an excellent way to remember.
precal
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|dw:1356824081954:dw|I always associated the laws with things I already know
itsjustme_lol
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i wish I could give u guys both medals :p i appreicate both!!
KingGeorge
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You're welcome.
precal
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np yw