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Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
08 Aug 2010, 03:10

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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).

Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
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? _________________

Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
08 Aug 2010, 07:41

1

This post received KUDOS

Expert's post

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). _________________

Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
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? _________________

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 \(3^{3x} = 90\) _________________

Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
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).

Re: Anything Wrong in this problem??? Can anyone dare to solve? [#permalink]
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).

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