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

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Joined: 02 Sep 2009
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If x and y are positive numbers such that x^y = y^x and y = 9x, then  [#permalink]

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15 Nov 2019, 01:44
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3
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Difficulty:

65% (hard)

Question Stats:

51% (02:31) correct 49% (02:31) wrong based on 51 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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Re: If x and y are positive numbers such that x^y = y^x and y = 9x, then  [#permalink]

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16 Nov 2019, 11:15
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Bunuel 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. $$\sqrt[4]{3}$$

$$x^y = y^x$$
—> $$x^{9x} = (9x)^x$$
—> $$x^{9x} = 9^x*x^x$$
—> $$x^{9x}/x^x = 9^x$$
—> $$x^{9x - x}= 9^x$$
—> $$x^{8x} = 9^x$$
—> $$(x^8)^x= 9^x$$
—> $$x^8 = 9 = 3^2$$
—> $$x = (3^2)^{1/8}$$
—> $$x = 3^{1/4}$$
—> $$x = \sqrt[4]{3}$$

IMO Option E

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

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16 Nov 2019, 13:19
Plug in y = 9x to get:

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

Now note x can be an exponent of both, so we can write it this way:

$$(x^9)^x = (9x)^x$$

If x cannot be 0, the only solution to this is to let the bases be the same.

$$x^9 = 9x$$

Again x cannot be 0 so we have $$x^8 = 9$$, therefore $$x =\sqrt[8]{9} = \sqrt[4]{3}$$

Ans: E

Bunuel 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. $$\sqrt[4]{3}$$

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

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19 Nov 2019, 04:31
Bunuel 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. $$\sqrt[4]{3}$$

$$x^y = y^x…y=9x…x^{9x}=(9x)^x$$
$$x^{9x}=(9x)^x…x^{9x}/x^x=9^x…x^{8x}=9^x$$
$$x^8x^x=9^x…x^8=9…\sqrt[8]{x^8}=\sqrt[8]{9}$$
$$x=9^{1/8}…x=(3^2)^{1/8}…x=3^{1/4}…x=\sqrt[4]{3}$$

Ans (E)
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Re: If x and y are positive numbers such that x^y = y^x and y = 9x, then  [#permalink]

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22 Nov 2019, 12:49
Bunuel 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. $$\sqrt[4]{3}$$

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We can substitute 9x for y in the first equation and obtain:

x^(9x) = (9x)^x

(x^x)^9 = 9^x * x^x

(x^x)^8 = 9^x

(x^8)^x = 9^x

Let’s take the xth root of each side. Notice that we can do this because we are given that x and y are positive integers:

x^8 = 9

x^8 = 3^2

x = 3^(2/8)

x = 3^(1/4)

So, x is the fourth root of 3.

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Re: If x and y are positive numbers such that x^y = y^x and y = 9x, then   [#permalink] 22 Nov 2019, 12:49
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