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Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application...
15 Nov 2019, 01:44
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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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. \(\sqrt[4]{3}\)
Are You Up For the Challenge: 700 Level Questions
A. \(9\)
B. \(\frac{1}{9}\)
C. \(\sqrt[9]{9}\)
D. \(\sqrt[3]{9}\)
E. \(\sqrt[4]{3}\)
Are You Up For the Challenge: 700 Level Questions
16 Nov 2019, 11:15
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Bunuel
\(x^y = y^x\)
—> \(x^{9x} = (9x)^x\)
—> \(x^{9x} = 9^x*x^x\)
—> \(x^{9x}/x^x = 9^x\)
—> \(x^{9x - x}= 9^x\)
—> \(x^{8x} = 9^x\)
—> \((x^8)^x= 9^x\)
—> \(x^8 = 9 = 3^2\)
—> \(x = (3^2)^{1/8}\)
—> \(x = 3^{1/4}\)
—> \(x = \sqrt[4]{3}\)
IMO Option E
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General Discussion
16 Nov 2019, 13:19
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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
\(x^{9x} = (9x)^x\)
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
