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If x and y are positive numbers such that x^y = y^x and y=9x, then the

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If x and y are positive numbers such that x^y = y^x and y=9x, then the  [#permalink]

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New post 19 Jan 2015, 06:25
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17
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A
B
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D
E

Difficulty:

  95% (hard)

Question Stats:

44% (02:37) correct 56% (02:27) wrong based on 215 sessions

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If x and y are positive numbers such that \(x^y = y^x\) and \(y=9x\), then the value of x is

a) \(9\)

b)\(\frac{1}{9}\)

c) \(\sqrt[9]{9}\)

d) \(\sqrt[3]{9}\)

e) \(\sqrt[4]{3}\)
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If x and y are positive numbers such that x^y = y^x and y=9x, then the  [#permalink]

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New post Updated on: 27 Oct 2017, 07:36
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\(x^y = y^x\)

\(x = y^{\frac{x}{y}} = (9x)^{\frac{x}{9x}}\)

\(x = 9^{\frac{1}{9}} * x^{\frac{1}{9}}\)

\(x^{\frac{8}{9}} = 3^{\frac{2}{9}}\)

\(x = 3^{\frac{2}{9} * \frac{9}{8}} = 3^{\frac{1}{4}}\)

Answer = E
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Originally posted by PareshGmat on 19 Jan 2015, 22:18.
Last edited by Mahmud6 on 27 Oct 2017, 07:36, edited 1 time in total.
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Re: If x and y are positive numbers such that x^y = y^x and y=9x, then the  [#permalink]

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New post 19 Jan 2015, 11:02
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Answer E.

Pluging the equation for y into the first one gives us

\(x^{9x}=9x^x\)

We can take the 8th radical and get

\(x^9=9x\)

Dividing by x gives

\(x^8=9\)

We don't have that in the answer choices, but we might notice the following relation:

\(x^8=9=3^2\)

So we can take the square root and get

\(x^4=3\)

And this leads to \(x=\sqrt[4]{3}\)
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If x and y are positive numbers such that x^y = y^x and y=9x, then the  [#permalink]

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New post 20 Jan 2015, 01:53
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1
\(x^y=y^x\)

y= \(\sqrt[x]{x^y}\)

now substitute the value of y=9x, we have

\(9x = x^{(9x/x)}\)

9x = x^9

x^8 = 9

or x= \(\sqrt[4]{3}\)
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If x and y are positive numbers such that x^y = y^x and y=9x, then the  [#permalink]

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New post 13 Jun 2017, 00:12
manpreetsingh86 wrote:
If x and y are positive numbers such that \(x^y = y^x\) and \(y=9x\), then the value of x is

a) \(9\)

b)\(\frac{1}{9}\)

c) \(\sqrt[9]{9}\)

d) \(\sqrt[3]{9}\)

e) \(\sqrt[4]{3}\)


\(x^y = y^x\) and \(y=9x\)
-> \(x^{9x}= (9x)^x\)
-> \(x^x (x^{8x} - 9^x) = 0\)
-> \(x^{8x} = 9^x = 3^{2x} =(\sqrt[2]{3})^{4x} = (\sqrt[4]{3})^{8x}\)
-> \(x = \sqrt[4]{3}\)

Answer E

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If x and y are positive numbers such that x^y = y^x and y=9x, then the  [#permalink]

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New post 31 Jul 2018, 11:49
manpreetsingh86 wrote:
If x and y are positive numbers such that \(x^y = y^x\) and \(y=9x\), then the value of x is

a) \(9\)

b)\(\frac{1}{9}\)

c) \(\sqrt[9]{9}\)

d) \(\sqrt[3]{9}\)

e) \(\sqrt[4]{3}\)


\(x^y = y^x\) and \(y=9x\)
\(x^9x = (9x)^x\)
\(x^9x = (9)^x * (x)^x\)
\(x^(9x-x) = 9^x\)
\(x^(8x) = (9)^x\)
\((x^8)^x = (9)^x\)
\(x^8 = 9\)
\(x = 9^(1/8)\)
\(x = 3^(2/8)\)
\(x = 3^(1/4)\)

Hence, Answer is E
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If x and y are positive numbers such that x^y = y^x and y=9x, then the &nbs [#permalink] 31 Jul 2018, 11:49
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