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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =

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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =  [#permalink]

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New post 04 Nov 2014, 09:43
1
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A
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D
E

Difficulty:

  5% (low)

Question Stats:

94% (01:25) correct 6% (01:36) wrong based on 167 sessions

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Re: (3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =  [#permalink]

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New post 10 Nov 2014, 02:21
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\((3^{5x} + 3^{5x} + 3^{5x})(4^{5x} + 4^{5x} + 4^{5x} + 4^{5x}) = 3 * 3^{5x} * 4 * 4^{5x}\)

\(= 3^{5x+1} * 4^{5x+1} = (3*4)^{5x+1} = 12^{5x+1}\)

Answer = A
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Re: (3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =  [#permalink]

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New post 04 Nov 2014, 10:25
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Factor out (3^(5x)) in the first parentheses and (4^(5x)) in the second parentheses. You get ((3^(5x)(3)) x ((4^5x)(4)). Since you have a common exponent (5x) multiply the basis (3^(5x))*(4^(5x)) = (12^(5x)). Noiw multiply the (3*4). So you get (12^(5x))*(12^1)), add common bases (12^(5x+1))
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Re: (3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =  [#permalink]

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New post 28 Dec 2014, 21:41
(1+1+1)*3^5x+(1+1+1+1)*4^5x=3^(5x+1)+4^(5x+1)=12^(5x+1)
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New post 25 Nov 2018, 05:28
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Re: (3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) =   [#permalink] 25 Nov 2018, 05:28
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