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Manager  Joined: 17 Mar 2010
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If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3?  [#permalink]

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Question Stats: 67% (02:18) correct 33% (02:43) wrong based on 445 sessions

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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

Originally posted by amitjash on 08 Aug 2010, 03:51.
Last edited by Bunuel on 12 Jan 2018, 11:42, edited 2 times in total.
Renamed the topic and edited the question.
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If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3?  [#permalink]

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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

$$3^{6x}=(3^{3x})^2=90^2=8,100$$ --> $$3^{3x}=90$$.

$$(3^{x-1})^3= 3^{3x-3}=\frac{3^{3x}}{3^3}=\frac{90}{27}=\frac{10}{3}$$.

Check Number Theory chapter of Math Book for exponents (link in my signature).
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If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3?  [#permalink]

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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
##### General Discussion
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Re: If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3?  [#permalink]

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

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

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

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

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

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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).

$$3^{6x}=(3^{3x})^2=90^2=8,100$$ --> $$3^{3x}=90$$.

$$(3^{x-1})^3= 3^{3x-3}=\frac{3^{3x}}{3^3}=\frac{90}{27}=\frac{10}{3}$$.

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

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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).

$$3^{6x}=(3^{3x})^2=90^2=8,100$$ --> $$3^{3x}=90$$.

$$(3^{x-1})^3= 3^{3x-3}=\frac{3^{3x}}{3^3}=\frac{90}{27}=\frac{10}{3}$$.

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

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$$3^{6x} = 8100$$

Square root both sides

$$3^{3x} = 90$$

Divide both sides by 27

$$\frac{3^{3x}}{27} = \frac{90}{27}$$

$$3^{3x-3} = \frac{10}{3}$$

$$[3^{(x-1)}]^3 = \frac{10}{3}$$

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

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_________________ Re: If 3^(6x) = 8,100, what is the value of [3^(x-1)]^3?   [#permalink] 12 Dec 2018, 07:42
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