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If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3?

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If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3? [#permalink] New post 08 Aug 2010, 02:51
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 May 2013, 07:57, 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] New post 08 Aug 2010, 03:10
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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] New post 08 Aug 2010, 07:18
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] New post 08 Aug 2010, 07:41
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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] New post 08 Aug 2010, 07:56
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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Re: Exponents [#permalink] New post 05 Dec 2010, 18:56
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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] New post 01 Jun 2013, 19:09
Bunuel wrote:
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).



Hi Bunnel,

Please point out the mistake in this approach. Why am I not getting correct answer by this method.

3^6x=8100=3^4*2^2*5^2
=> 6x=4, as all nos are primes
=> x= 2/3

Now, 3^(3*(x-1)) = 3^(3*(-1/3)) = 3^(-1) = 1/3.

Still based on denominator I chose D.
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Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink] New post 02 Jun 2013, 03:13
Expert's post
cumulonimbus wrote:
Bunuel wrote:
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).



Hi Bunnel,

Please point out the mistake in this approach. Why am I not getting correct answer by this method.

3^6x=8100=3^4*2^2*5^2
=> 6x=4, as all nos are primes
=> x= 2/3

Now, 3^(3*(x-1)) = 3^(3*(-1/3)) = 3^(-1) = 1/3.

Still based on denominator I chose D.


From 3^6x=8100=3^4*2^2*5^2 we cannot say that 6x=4 --> 3^4=8100=3^4*2^2*5^2 --> 1=2^2*5^2 which is not correct.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3? [#permalink] New post 26 Feb 2014, 22:38
3^6x = 8100

3^3x . 3^3x = 90 . 90

3^3x = 90

Divide both sides by 27

3^(3x-3) = 90/27

\frac{{3^(x-1)}^3[}{fraction] = [fraction]10/3} = Answer = C
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Re: If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3?   [#permalink] 26 Feb 2014, 22:38
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