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If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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10 Jul 2008, 19:05
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If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x? A. 9 B. 6 C. 3 D. 6 E. 9
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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04 Jan 2013, 06:08
xALIx wrote: If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x?
A. 9 B. 6 C. 3 D. 6 E. 9 \(27^{4x + 2} * 162^{2x} * 36^x * 9^{62x} = 1\) \(3^{3*(4x + 2)} * (2^{2x}*81^{2x}) * (4^{x}*9^x) * 3^{2(62x)} = 1\) \(3^{3*(4x + 2)} * (2^{2x}*3^{8x}) * (2^{2x} *3^{2x})* 3^{2(62x)} = 1\) > \(2^{2x}*2^{2x}=1\). So, we have that: \(3^{3*(4x + 2)8x+2x+2(62x)} = 1\) \(3^{2x+18}=1\) > \(2x+18=0\) > \(x=9\). Answer: A.
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Re: PS  Powers, what is the value of x?
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10 Jul 2008, 19:14
xALIx wrote: If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x?
A) 9 B) 6 C) 3 D) 6 E) 9 (3^3)^(4x + 2) * (3^4 * 2)^2x * (3^2 * 2^2) ^x * (3^2)^(62x) = 1 the 2 ^ 2x from second term and 2^2x from the third term gets canclled giving us all all terms with base 3. 3^ (2x + 18) = 1 2x + 18 must be 0 in order to satisfy this x= 9 A



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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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14 Nov 2012, 13:03
This is the last part of the answer from the CAT explanation.
2^0 × 3^2x + 18 = 1 3^2x + 18 = 1 3^2x + 18 = 3^0 2x + 18 = 0 2x = 18 x = 9
where does the 3^0 comes from?



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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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14 Nov 2012, 13:06
3^0 is substituted for 1 to show both sides of the equation with the same base
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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04 Jan 2013, 05:10
Can anyone explain how the 18 comes into play?
Also 2^0 in the second explanation.
Thanks.



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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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16 May 2013, 12:38
Quick question: How I am I supposed to know that 162 equals 2*3^4 ??? Is there a calculation to it or is 81= 3^4 common knowledge? thanks



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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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16 May 2013, 23:58
NicolasFSS wrote: Quick question: How I am I supposed to know that 162 equals 2*3^4 ??? Is there a calculation to it or is 81= 3^4 common knowledge? thanks If you don't know that 81=3^4 you can do the following: 162=2*81=2*9*9=2*3^2*3^2=2*3^4. I'd advice to know the powers of 2 till 2^10 and the powers of 3 till 3^5.
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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27 Feb 2014, 01:03
Bunuel wrote: xALIx wrote: If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x?
A. 9 B. 6 C. 3 D. 6 E. 9 \(27^{4x + 2} * 162^{2x} * 36^x * 9^{62x} = 1\) \(3^{3*(4x + 2)} * (2^{2x}*81^{2x}) * (4^{x}*9^x) * 3^{2(62x)} = 1\) \(3^{3*(4x + 2)} * (2^{2x}*3^{8x}) * (2^{2x} *3^{2x})* 3^{2(62x)} = 1\) > \(2^{2x}*2^{2x}=1\). So, we have that: \(3^{3*(4x + 2)8x+2x+2(62x)} = 1\) \(3^{2x+18}=1\) > \(2x+18=0\) > \(x=9\). Answer: A. I did in the same way, however taking more time Any shorter approach please?
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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27 Feb 2014, 05:30
PareshGmat wrote: Bunuel wrote: xALIx wrote: If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x?
A. 9 B. 6 C. 3 D. 6 E. 9 \(27^{4x + 2} * 162^{2x} * 36^x * 9^{62x} = 1\) \(3^{3*(4x + 2)} * (2^{2x}*81^{2x}) * (4^{x}*9^x) * 3^{2(62x)} = 1\) \(3^{3*(4x + 2)} * (2^{2x}*3^{8x}) * (2^{2x} *3^{2x})* 3^{2(62x)} = 1\) > \(2^{2x}*2^{2x}=1\). So, we have that: \(3^{3*(4x + 2)8x+2x+2(62x)} = 1\) \(3^{2x+18}=1\) > \(2x+18=0\) > \(x=9\). Answer: A. I did in the same way, however taking more time Any shorter approach please? Not every question has a silver bullet approach.
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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08 Mar 2014, 20:38
Option A. All terms are variants of power of 3. We reduce each term to some power of 3 and after cancellation get 3^(2x+18)=1 2x+18=0 X=9
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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23 Aug 2015, 02:37
Hi any shorter route to solution? this is a classical manhattan cat question  simple concept, a lot of iterations. gmac would unlikely post something like this, wouldnt it? looking forward to any shortcut
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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23 Aug 2015, 05:49
xALIx wrote: If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x?
A. 9 B. 6 C. 3 D. 6 E. 9 You have only two prime factors, 2 and 3, involved in this question of exponents. 162=2*3^4 36 =2^2 * 3^2 Therefore, 2^2x and 2^2x get cancelled. Now, check only exponents of 3 3^12x * 3^6 * 3^8x * 3^2x * 3^12 * 3^4x=1 3^2x * 3^18 = 3^0 Implies 2x+18 = 0 x = 9



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Technique: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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08 Oct 2016, 05:27
Hi everyone  Thank you all for your posts.. I thought I would try do my bit Powers and bases and rate questions have been my nemisis when it has come to pushing the upper ends of quant scores  mainly as I kept getting these wrong! .. I would lose track of the calculations in trying to keep the scientific notation. It was costing me 12 questions a test and really frustrating me but most importantly was costing time. The below technique is one which has turned these problems on their head for me ... i hope its useful It is not perfect but its very helpful for these types of questions, especially if like me you fight the constant battle against the demon of attention to detail (and his sidekick  messy writing) My method for base questions is analyse vertically in form Base  Power  Variable(s)  Exponent. Using this, this question came together in around 1m50 while I reckon it would have been a 3m+ expedition prior to this method. Step 1 .... quick scan reveals that this is likely to be a base question. Sketch quick table and then do the factorisation one by one. 27  3^3 162 = 81 x 2 = 3^4 and 2^1 36 = 9 x 4 = 3^2 and 2^2 9 = 3^2 Powers of 2 cancel Base 3 / Exponent = 12x – 8x + 2x – 4x + 12 + 6 = 18 +2x 3^(18+2x) = 3^0 X= 9



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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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21 Oct 2016, 06:53
xALIx wrote: If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x?
A. 9 B. 6 C. 3 D. 6 E. 9 needs to be used math function on the forum...it gets difficult to understand what is what... 27 = 3^3 27^(4x + 2) = 3^(12x+6) 162 = 3^4 * 2 162^2x = 3^(8x) * 2^(2x) 36^x = 3^2x * 2^2x 9^(6 – 2x) = 3^(124x) now. 3^(12x+6) * 3^(8x) * 2^(2x) * 3^2x * 2^2x * 3^(124x) powers of 2 cancel, and we have 2^0, which is equal to 1. rest, we can rewrite as 3 to a long power: 12x+68x+2x+124x=2x+18. since 3^(2x+18) = 1, it must be true that 2x+18=0. x=9
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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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19 Apr 2017, 23:00
xALIx wrote: If 27^(4x + 2) × 162^2x × 36^x × 9^(6 – 2x) = 1, then what is the value of x?
A. 9 B. 6 C. 3 D. 6 E. 9 No shortcuts in this problem best to just think fast and work hard to force out the answer 74x + 2 × 1622x × 36x × 96 – 2x = 1 (33)4x + 2 × (2 × 34)2x × (22 × 32)x × (32)6 – 2x = 1 312x + 6 × 22x × 38x × 22x × 32x × 312 – 4x = 1 22x + 2x × 312x + 6 – 8x + 2x + 12 – 4x = 1 20 × 3 2x + 18 = 1 3 2x + 18 = 1 Any number to the zero power equals 1 so set 3's algebraic expression exponent equal to 0 2x + 18 = 0 (9) The correct answer is A.



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Re: If 27^(4x + 2) 162^2x 36^x 9^(6 2x) = 1, then what is
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