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If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3? [#permalink]
 08 Aug 2010, 03:51
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If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3? A. 90 B. 30 C. 10 D. 10/3 E. 10/9
Last edited by Bunuel on 16 May 2013, 08:57, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
08 Aug 2010, 04:10
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amitjash wrote: If 3^6x = 8,100, what is the value of (3^x – 1)^3 ?
A. 90
B. 30
C. 10
D. 10/3
E. 10/9 Please when posting such questions make sure that it's not ambiguous. Correct question is: If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3? Or it can be written using the formating as: If 3^{6x}=8,100, what is the value of (3^{x-1})^3? (It's not hard at all). 3^{6x}=(3^{3x})^2=90^2=8,100 --> 3^{3x}=90. (3^{x-1})^3= 3^{3x-3}=\frac{3^{3x}}{3^3}=\frac{90}{27}=\frac{10}{3}. Answer: D. Check Number Theory chapter of Math Book for exponents (link in my signature).
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Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
08 Aug 2010, 08:18
Excellent. No a side note, Bunuel, I started approaching this using the "power of prime in a number" - I had seen the link somewhere which had the formula N/p + N/p^2+N/p^3... until N>p^x and thought of equating that to 6x and solve thus. I know yours is much simpler. However would that approach have worked? Is that formula exactly saying what is the highest power of the prime in the number?
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Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
08 Aug 2010, 08:41
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mainhoon wrote: Excellent. No a side note, Bunuel, I started approaching this using the "power of prime in a number" - I had seen the link somewhere which had the formula N/p + N/p^2+N/p^3... until N>p^x and thought of equating that to 6x and solve thus. I know yours is much simpler. However would that approach have worked? Is that formula exactly saying what is the highest power of the prime in the number? No need to complicate simple questions. The formula is correct ( everything-about-factorials-on-the-gmat-85592.html) but it has nothing to do with this problem, (highest power of 3 in 81,000 won't be equal to 6x, because 3^(6x)=81,000=2^m*3^n*5^k, so as 81,000 has other factors than 3 in it then 6x won't be an integer at all).
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Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
08 Aug 2010, 08:56
When I think some more, I don't think we can use that formula. Isn't that formula for factorials - (a) N! not N and (b) as you rightly pointed out it would result in a fraction for 6x, not integer. For 3^(3x) = 90, if I wanted to solve it, it is clear than x is fractional. So is the answer basically taking logarithms? Any other way?
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Q11: If 3^6x = 8,100, what is the value of 3^((x – 1)3) ? 3 A. 90 B. 30 C. 10 D. 10/3 E. 10/9
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ajit257 wrote: Q11:
If 3^6x = 8,100, what is the value of 3^((x – 1)3) ?
3
A. 90
B. 30
C. 10
D. 10/3
E. 10/9 In a question such as this where you have 3^{6x} = 8100 where 8100 is not a perfect sixth power but the options do not have irrational numbers, you should immediately go and analyze what is asked. You will not need to solve for x. You will end up using either 3^{6x} = 8100 or 3^{3x} = 90
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Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
07 Dec 2010, 14:50
Q11:
If 3^6x = 8,100, what is the value of 3^((x – 1)3) ?
3
A. 90
B. 30
C. 10
D. 10/3
E. 10/9
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Re: Anything Wrong in this problem??? Can anyone dare to solve?
[#permalink]
07 Dec 2010, 14:50
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