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Re: If 8^0.5y 3^0.75x = 12^n then what is the value of x?? 1) n [#permalink]
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LM wrote:
\((2^(3*y/2))*(3^(3*x/4))=(2^6)*(3^3)\)

\(3*y/2=6\)
\(y=4\)
\(3*x/4=3\)
\(x=4\)

Therefore "A"

Buenel, what's the mistake in my approach?


(1) \(n=3\) --> \(2^{\frac{3y}{2}}*3^{\frac{3x}{4}}=2^6*3^3\) --> now you could equate the powers if you knew that both \(x\) and \(y\) are integers, but we don't know that.

For example if \(y=0\) --> \(3^{\frac{3x}{4}}=2^6*3^3\) and in this case \(x\) will be some irrational number.
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Re: If 8^0.5y 3^0.75x = 12^n then what is the value of x?? 1) n [#permalink]
Bunuel wrote:
surendar26 wrote:
If 8^0.5y 3^0.75x = 12^n then what is the value of x??

1) n = 3.
2) Both x and y are natural numbers.


Not a GMAT question.

\(8^{\frac{y}{2}}*3^{\frac{3x}{4}}=12^n\) --> \(2^{\frac{3y}{2}}*3^{\frac{3x}{4}}=2^{2n}*3^n\) --> \(2^{\frac{3y}{2}-2n}*3^{\frac{3x}{4}-n}=1\). Now, either powers of 2 and 3 are both zero or these two multiples are reciprocals of each other and in this case powers of 2 and/or 3 must be some irrational numbers.

(1) n = 3.
(2) Both x and y are natural numbers.

Each statement alone is not sufficient, as we can not decide which case we have.

(1)+(2) As all variables are integers then the powers of 2 and 3 can not be irrational numbers thus they must equal to zero: \(\frac{3x}{4}-n=\frac{3x}{4}-3=0\) --> \(x=4\). Sufficient.

Answer: C.




Hello Bunnel,

You have mentioned this is not a GMAT Question..
May I know the reason.
Because I find this pattern similar to a GMAT Question.
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Re: If 8^0.5y 3^0.75x = 12^n then what is the value of x?? 1) n [#permalink]
Bunuel, I couldn't understand your statement "you could equate the powers if you knew that both X and Y are integers, but we don't know that ".
RHS and LHS are both represented in their respective prime factors. So what's the concept behind not equating the powers in this case?
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Re: If 8^0.5y 3^0.75x = 12^n then what is the value of x?? 1) n [#permalink]
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vjsharma25 wrote:
Bunuel, I couldn't understand your statement "you could equate the powers if you knew that both X and Y are integers, but we don't know that ".
RHS and LHS are both represented in their respective prime factors. So what's the concept behind not equating the powers in this case?


(1) \(n=3\) --> \(2^{\frac{3y}{2}}*3^{\frac{3x}{4}}=2^6*3^3\) --> now you could equate the powers if you knew that both \(x\) and \(y\) are integers, but we don't know that.

For example if \(y=0\) --> \(3^{\frac{3x}{4}}=2^6*3^3\) and in this case \(x\) will be some irrational number. Basically \(2^{\frac{3y}{2}}*3^{\frac{3x}{4}}=2^6*3^3\) has infinitely many solutions for \(x\) and \(y\): for any \(y\) there will exist some \(x\) to satisfy this equation. If we were told that both are integers then these equation would have only one integer solution: \(x=4\) and \(y=4\)

Similar question: disagree-with-oa-106047.html
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Re: If 8^0.5y 3^0.75x = 12^n then what is the value of x?? 1) n [#permalink]
So important takeaway from this question is you can not equate the powers in an equality if you do not know whether the variables making the powers are integers or not.
Can we say that?
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Re: If 8^0.5y 3^0.75x = 12^n then what is the value of x?? 1) n [#permalink]
surendar26 wrote:
If 8^0.5y 3^0.75x = 12^n then what is the value of x??

1) n = 3.
2) Both x and y are natural numbers.


question is not properly presented and i got it wrong as i assumed some other expression :(.

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