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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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Senior Manager
Joined: 13 Jun 2013
Posts: 277
If x and y are positive numbers such that x^y = y^x and y=9x, then the  [#permalink]

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19 Jan 2015, 06:25
2
17
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Difficulty:

95% (hard)

Question Stats:

47% (02:15) correct 53% (02:26) wrong based on 207 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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Updated on: 27 Oct 2017, 07:36
7
$$x^y = y^x$$

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

$$x = 9^{\frac{1}{9}} * x^{\frac{1}{9}}$$

$$x^{\frac{8}{9}} = 3^{\frac{2}{9}}$$

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

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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.
Removing Typo
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GMAT 1: 770 Q50 V45
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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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19 Jan 2015, 11:02
3
3

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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20 Jan 2015, 01:53
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$$x^y=y^x$$

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

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

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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