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Sub 505 (Easy)|   Exponents|      
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Bunuel
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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 04 Nov 2014, 09:43
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A
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86% (01:21) correct 14% (01:59) wrong based on 369 sessions

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Tough and Tricky questions: Exponents.



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

A. 12^(5x + 1)
B. 3^(15x) + 4^(20x)
C. 25^(5x)
D. 7^(35x)
E. 25^(5x+1)

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A
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PareshGmat
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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 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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bankerboy30
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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 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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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 28 Dec 2014, 21:41
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(1+1+1)*3^5x+(1+1+1+1)*4^5x=3^(5x+1)+4^(5x+1)=12^(5x+1)
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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 20 Apr 2020, 07:40
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\(\)
Bunuel

Tough and Tricky questions: Exponents.



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

A. 12^(5x + 1)
B. 3^(15x) + 4^(20x)
C. 25^(5x)
D. 7^(35x)
E. 25^(5x+1)

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We cannot add exponents even if bases are same, so have to factor out.
First part is basically \(3*3^{5x}\)and second part is \(4*4^{5x}\)

Now that we have brought the equation to a point where laws of exponents work we can start solving.

\(3^1*3^{5x} = 3^{5x+1}\) (Remember adding exponents when base is same and it is being multiplied. Same with second part \(4^{1}*4^{5x}=4^{5x+1} \).

Final = \(3^{5x+1} * 4^{5x+1}\) . Applying distribution law. \((3*4)^{5x+1} = 12^{5x+1}\)

Thus answer A.
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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 25 Mar 2021, 04:48
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Substitute x=1/5 in the question
(3+3+3)(4+4+4+4)=9*16=144

Substitute x=1/5 in the answers to find the match.

A)12^(5x+1)=>12^(5(1/5)+1)=12^(2)=144 -Match
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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 06 Mar 2023, 19:00
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Remember that when you multiply different bases raised to the SAME exponent, the product is simply the product of the bases raised to their common exponent.

(3^5x +3^5x +3^5x)(4^5x +4^5x +4^5x +4^5x)= 3^5x(1 + 1 + 1) × 4^5x(1 + 1 + 1 + 1) =
3(3^5x) × 4(4^5x) = 3^5x+1 ×4^5x+1 =
(3 × 4)^5x+1 = 12^5x+1

The correct answer is A.
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(3^5x + 3^5x + 3^5x)(4^5x + 4^5x + 4^5x + 4^5x) = 11 Oct 2024, 16:49
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