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December 25th, 2013, 10:18 AM   #1
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Exponents

Q: The value of (x,y) if and is:

(a)

(b)

(c)

(d)

ANS: Can this problem be solved algebraically, without plugging in the answer options into the question stem? Request an algebraic solution.
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December 25th, 2013, 12:20 PM   #2
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As x = y, y = x^(2/3), and so x^y = (x^(2/3))^x = x^((2/3)x).
Hence y = (2/3)x, and so x = (2/3)x.
Hence x = (3/2) = 27/8, and so y = (2/3)(27/ = 9/4.
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December 26th, 2013, 09:36 PM   #3
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Re:

Quote:
Originally Posted by skipjack
As x = y, y = x^(2/3), and so[color=#FF0000] x^y = (x^(2/3))^x = x^((2/3)x)[/color].
Hence y = (2/3)x, and so x = (2/3)x.
Hence x = (3/2) = 27/8, and so y = (2/3)(27/ = 9/4.
I don't understand the part in bold skipjack. Shouldn't it be x^y = x^(x^(2/3)). How does this equal x^((2/3)x)?
My solution looks like this
x^y = x^(x^(2/3)) = y^x = (x^(2/3))^x = x^(2x/3)
so x^(x^(2/3)) = x^(2x/3)
so x^(2/3) = 2x/3
so x = 27/8 and y = 9/4

Please clarify. Thanks
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December 27th, 2013, 06:26 PM   #4
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The equation you asked about in my solution is simply a shortened version of the first line in your solution.
I substituted for y only in the right-hand side of the equation x^y = y^x.
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December 28th, 2013, 10:33 AM   #5
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Re: Exponents

Hi All !

I have made a solution of my own. Please inform whether it is correct or not.

Since so

And so

Now, since the base is same on both sides = y, we can equate the exponents:



or,

or,

so,

And, since





Now, please inform whether my solution was correct or wrong! Please inform soon.
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December 28th, 2013, 03:06 PM   #6
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Re: Exponents

That's correct.
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December 28th, 2013, 10:55 PM   #7
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There is another solution, but it isn't listed as a possibility. What is it?
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