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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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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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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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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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Bunuel
amitjash
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.
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\(3^{6x} = 8100\)

Square root both sides

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

Divide both sides by 27

\(\frac{3^{3x}}{27} = \frac{90}{27}\)

\(3^{3x-3} = \frac{10}{3}\)

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

Answer = D
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Let \((3^{x - 1})^3\) = p

=> \((3^{x})^3 * (3^{-1})^3\) = p

=> \(\frac{(3^{3x})}{ 3^{3}}\) = p

=> Squaring both the sides: \(\frac{((3^{3x})^2)}{((3^{3})^2)} = p^2 \)

=> \(\frac{(3^{6x})}{3^{6}} = p^2\)

=> \(\frac{8100 }{ 3^6} = p^2\)

=> \(\frac{100 }{ 9} = p^2\)

=> p = \(\frac{10}{3}\)

Answer D
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3^6x = 8100
3^6x = 3^ 4 * 100

3^(6x-4) = 100
3 ^ (3x-2) = 10

We are asked for 3^ (3x-3). Therefore answer is 10/3.
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Bunuel
amitjash
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

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

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

Answer: D.

Check Number Theory chapter of Math Book for exponents (link in my signature).

Bunuel - This is just such an elegant solution. I didnt even think of just separting the chunk and solving for 3^3x as a chunk. So much value in your posts. Thanks for all you do.
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