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If x^(1-x)=4, what is the value of
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13 Feb 2019, 19:28
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31% (02:45) correct
If \({x^{1 - x}} = 4\), what is the value of \({x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\) ?
\(\left( A \right)\,\sqrt 2\)
\(\left( B \right)\,1\)
\(\left( C \right)\,\sqrt {0.5}\)
\(\left( D \right)\,\sqrt {0.25}\)
\(\left( E \right)\,\sqrt {0.125}\)
\(\left( A \right)\,\sqrt 2\)
\(\left( B \right)\,1\)
\(\left( C \right)\,\sqrt {0.5}\)
\(\left( D \right)\,\sqrt {0.25}\)
\(\left( E \right)\,\sqrt {0.125}\)
Director
Joined: 09 Mar 2018
Posts: 782
Given Kudos: 123
Location: India
Re: If x^(1-x)=4, what is the value of
[#permalink]
13 Feb 2019, 20:18
1
Kudos
fskilnik wrote:
GMATH practice exercise (Quant Class 12)
If \({x^{1 - x}} = 4\), what is the value of \({x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\) ?
\(\left( A \right)\,\sqrt 2\)
\(\left( B \right)\,1\)
\(\left( C \right)\,\sqrt {0.5}\)
\(\left( D \right)\,\sqrt {0.25}\)
\(\left( E \right)\,\sqrt {0.125}\)
If \({x^{1 - x}} = 4\), what is the value of \({x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\) ?
\(\left( A \right)\,\sqrt 2\)
\(\left( B \right)\,1\)
\(\left( C \right)\,\sqrt {0.5}\)
\(\left( D \right)\,\sqrt {0.25}\)
\(\left( E \right)\,\sqrt {0.125}\)
I was so close on this one.( a great question of manipulation)
x = 4 * \(x^x\)
\({x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\)
\(x^{\frac{x}{4} - \frac{1}{4}}\)
x^{x/4} / x^{1/4}
basically 1/4 is nothing but four root of a number
\(4\sqrt{x^x}\) / \(4\sqrt{x}\)
The above expression will become \(4\sqrt{1/4}\)
which will become \(\sqrt{0.5}\)
Re: If x^(1-x)=4, what is the value of
[#permalink]
14 Feb 2019, 06:08
1
Kudos
Expert Reply
fskilnik wrote:
GMATH practice exercise (Quant Class 12)
If \({x^{1 - x}} = 4\), what is the value of \({x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\) ?
\(\left( A \right)\,\sqrt 2\)
\(\left( B \right)\,1\)
\(\left( C \right)\,\sqrt {0.5}\)
\(\left( D \right)\,\sqrt {0.25}\)
\(\left( E \right)\,\sqrt {0.125}\)
If \({x^{1 - x}} = 4\), what is the value of \({x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\) ?
\(\left( A \right)\,\sqrt 2\)
\(\left( B \right)\,1\)
\(\left( C \right)\,\sqrt {0.5}\)
\(\left( D \right)\,\sqrt {0.25}\)
\(\left( E \right)\,\sqrt {0.125}\)
Thanks for your contribution (kudos!) and for the kind words, KanishkM !
Let me present our "official solution":
\({x^{1 - x}} = 4\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^{ - \left( {1 - x} \right)}} = {4^{ - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^{x - 1}} = {1 \over 4}\,\,\,\,\,\,\,\left( * \right)\)
\(?\,\, = \,\,{x^{{x^x}\,\, - \,\,{{{x^x}} \over x}}}\,\, = \,\,{x^A}\,\,\,\,\,\,\,\,\,\,\,\,\left( {A = {x^x}\,\, - \,\,{{{x^x}} \over x}} \right)\)
\(A\,\, = \,\,{x^x} - {x^{x - 1}}\,\, = \,\,{x^{x - 1}}\left( {x - 1} \right)\,\,\mathop = \limits^{\left( * \right)} \,\,{{x - 1} \over 4}\)
\(?\,\, = \,\,{x^{{{x - 1} \over 4}}}\,\, = \,\,\root 4 \of {{x^{x - 1}}} \,\,\mathop = \limits^{\left( * \right)} \,\,\root 4 \of {{1 \over 4}} \,\, = \,\,\root {2 \cdot 2} \of {{{\left( {{1 \over 2}} \right)}^2}} \,\, = \,\,\sqrt {0.5}\)
The correct answer is (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.