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 December 25th, 2013, 10:18 AM #1 Member   Joined: Dec 2013 Posts: 86 Thanks: 1 Exponents Q: The value of (x,y) if $x^y=y^x$ and $x^2=y^3$is: (a)$\frac{27}{8} , \frac{9}{4}$ (b)$\frac{9}{8} , \frac{27}{4}$ (c)$\frac{8}{27}, \frac{4}{9}$ (d)$\frac{8}{9} , \frac{4}{27}$ ANS:$(a).$ Can this problem be solved algebraically, without plugging in the answer options into the question stem? Request an algebraic solution.
 December 25th, 2013, 12:20 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,417 Thanks: 1462 As x² = y³, y = x^(2/3), and so x^y = (x^(2/3))^x = x^((2/3)x). Hence y = (2/3)x, and so x² = (2/3)³x³. Hence x = (3/2)³ = 27/8, and so y = (2/3)(27/ = 9/4.
December 26th, 2013, 09:36 PM   #3
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Quote:
 Originally Posted by skipjack As x² = y³, 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 x² = (2/3)³x³. Hence x = (3/2)³ = 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

 December 27th, 2013, 06:26 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,417 Thanks: 1462 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.
 December 28th, 2013, 10:33 AM #5 Member   Joined: Dec 2013 Posts: 86 Thanks: 1 Re: Exponents Hi All ! I have made a solution of my own. Please inform whether it is correct or not. Since $x^2=y^3$ so $x=y^{\frac{3}{2}}$ And $x^y=y^x$ so $y^{\frac{3y}{2}}=y^{y}^{\frac{3}{2}}$ Now, since the base is same on both sides = y, we can equate the exponents: $\frac{3y}{2}=y^{\frac{3}{2}}$ or, $\frac{3}{2}=y^{\frac{3}{2}-1}$ or, $\frac{3}{2}=y^{\frac{1}{2}}$ so, $y=\frac{9}{4}$ And, since $x=y^{\frac{3}{2}}$ $x=(\frac{9}{4})^{\frac{3}{2}}$ $x=\frac{27}{8}$ Now, please inform whether my solution was correct or wrong! Please inform soon.
 December 28th, 2013, 03:06 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,692 Thanks: 976 Math Focus: Elementary mathematics and beyond Re: Exponents That's correct.
 December 28th, 2013, 10:55 PM #7 Global Moderator   Joined: Dec 2006 Posts: 18,417 Thanks: 1462 There is another solution, but it isn't listed as a possibility. What is it?

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