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21 Feb 2023, 13:13
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If \(x^x=\sqrt{\frac{\sqrt{2}}{2}}\), which of the following could be a value of \(x\)?
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\frac{1}{2}\)
E. \(\sqrt{2}\)
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\frac{1}{2}\)
E. \(\sqrt{2}\)
21 Feb 2023, 13:13
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Official Solution:
If \(x^x=\sqrt{\frac{\sqrt{2}}{2}}\), which of the following could be a value of \(x\)?
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\frac{1}{2}\)
E. \(\sqrt{2}\)
We need to simplify the expression on the right-hand side of the equation so that it can be written in terms of the same base and exponent as \(x^x\).
\(\sqrt{\frac{\sqrt{2}}{2}}=\)
\(=\sqrt{\frac{1}{\sqrt{2}}}=\)
\(=\sqrt{(\frac{1}{2})^{\frac{1}{2}}}=\)
\(=(\frac{1}{2})^{\frac{1}{4}}=\)
Take to the above expression the fourth power and apply exponentiation to the base, \(\frac{1}{2}\):
\(=((\frac{1}{2})^{\frac{1}{4}})^4=\)
\(=((\frac{1}{2})^4)^{\frac{1}{4}}=\)
\(=(\frac{1}{16})^{\frac{1}{4}}\)
Now, to compensate the previous operation, take the fourth root, but this time apply exponentiation to the exponent, \(\frac{1}{4}\):
\(=((\frac{1}{16})^{\frac{1}{4}})^{\frac{1}{4}}=\)
\(=(\frac{1}{16})^{\frac{1}{16}}\)
Therefor, \(x=\frac{1}{16}\)
Else, after obtaining the expression \(x^x=(\frac{1}{2})^{\frac{1}{4}}\), we can substitute each of the given options for \(x\) to determine which one satisfies the equation.
Option A is a valid solution, as \((\frac{1}{16})^{(\frac{1}{16})}=((\frac{1}{16})^{(\frac{1}{4})})^{(\frac{1}{4})}=(\frac{1}{2})^{\frac{1}{4}}\)
Answer: A
If \(x^x=\sqrt{\frac{\sqrt{2}}{2}}\), which of the following could be a value of \(x\)?
A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\frac{1}{2}\)
E. \(\sqrt{2}\)
We need to simplify the expression on the right-hand side of the equation so that it can be written in terms of the same base and exponent as \(x^x\).
\(\sqrt{\frac{\sqrt{2}}{2}}=\)
\(=\sqrt{\frac{1}{\sqrt{2}}}=\)
\(=\sqrt{(\frac{1}{2})^{\frac{1}{2}}}=\)
\(=(\frac{1}{2})^{\frac{1}{4}}=\)
Take to the above expression the fourth power and apply exponentiation to the base, \(\frac{1}{2}\):
\(=((\frac{1}{2})^{\frac{1}{4}})^4=\)
\(=((\frac{1}{2})^4)^{\frac{1}{4}}=\)
\(=(\frac{1}{16})^{\frac{1}{4}}\)
Now, to compensate the previous operation, take the fourth root, but this time apply exponentiation to the exponent, \(\frac{1}{4}\):
\(=((\frac{1}{16})^{\frac{1}{4}})^{\frac{1}{4}}=\)
\(=(\frac{1}{16})^{\frac{1}{16}}\)
Therefor, \(x=\frac{1}{16}\)
Else, after obtaining the expression \(x^x=(\frac{1}{2})^{\frac{1}{4}}\), we can substitute each of the given options for \(x\) to determine which one satisfies the equation.
Option A is a valid solution, as \((\frac{1}{16})^{(\frac{1}{16})}=((\frac{1}{16})^{(\frac{1}{4})})^{(\frac{1}{4})}=(\frac{1}{2})^{\frac{1}{4}}\)
Answer: A
17 Feb 2025, 22:04
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Another way is to keep on squaring both the sides and then using options to get to the correct choice
\(x^x=\sqrt{\frac{\sqrt{2}}{2}}\)
\(x^{2x}=\frac{1}{\sqrt{2}}\)
\(x^{4x}=\frac{1}{2}\)
This is now easy to validate against option choices, let's try A,
\(x^{4x}={(\frac{1}{16}})^{\frac{1}{4}}={(\frac{1}{4})}^{\frac{1}{2}}=\frac{1}{2}\)
Ans: A
\(x^x=\sqrt{\frac{\sqrt{2}}{2}}\)
\(x^{2x}=\frac{1}{\sqrt{2}}\)
\(x^{4x}=\frac{1}{2}\)
This is now easy to validate against option choices, let's try A,
\(x^{4x}={(\frac{1}{16}})^{\frac{1}{4}}={(\frac{1}{4})}^{\frac{1}{2}}=\frac{1}{2}\)
Ans: A
