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

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New post 15 Nov 2019, 01:44
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Question Stats:

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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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New post 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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New post 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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New post 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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New post 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.

Answer: E
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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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