Nilai x yang memenuhi \left(\sqrt x\right)^x" data-mathml="«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msup»«mi»x«/mi»«msqrt»«mi»x«/mi»«/msqrt»«/msup»«mo»§#62;«/mo»«msup»«mfenced»«msqrt»«mi»x«/mi»«/msqrt»«/mfenced»«mi»x«/mi»«/msup»«/math»" role="math" src="data:image/png;base64,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" style="max-width: none;">adalah ...

Pembahasan

Ingat! Pertidaksamaan eksponen bentuk a f ( x ) > a g ( x ) mempunyai penyelesaian: f ( x ) > g ( x ) jika a > 1 f ( x ) < g ( x ) jika 0 < a < 1 Diketahui \left(\sqrt x\right)^x" data-mathml="«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msup»«mi»x«/mi»«msqrt»«mi»x«/mi»«/msqrt»«/msup»«mo»§#62;«/mo»«msup»«mfenced»«msqrt»«mi»x«/mi»«/msqrt»«/mfenced»«mi»x«/mi»«/msup»«/math»" role="math" src="data:image/png;base64,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" style="max-width: none;"> ( i ) Agar terdefinisi maka x ≥ 0 ( ii ) Jika x > 1 maka &\left(\sqrt x\right)^x\\x^\sqrt x&>&x^{\frac12x}\\\sqrt x&>&\frac12x\\x&>&\left(\frac12x\right)^2\\x&>&\frac14x^2\\-\frac14x^2+x&>&0\\x^2-4x& Diperoleh himpunan penyelesaiannya adalah { x > 1 } ∩ { 0 < x < 4 } = { 1 < x < 4 } ( iii ) Jika 0 < x < 1 maka &\left(\sqrt x\right)^x\\x^\sqrt x&>&x^{\frac12x}\\\sqrt x& &0\\x\left(x-4\right)&>&0\\x& 4\end{array}" data-mathml="«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center 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mathvariant=¨normal¨»x«/mi»«mo»§#62;«/mo»«mn»4«/mn»«/mtd»«/mtr»«/mtable»«/math»" role="math" 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style="max-width: none;"> Diperolehhimpunan penyelesaian adalah { 0 < x < 1 } ∩ { x < 0 atau x > 4 } = { } Daerah penyelesaian dari pertidaksamaan di atas adalah irisan himpunan penyelesaian (i), (ii), dan (iii). { x ≥ 0 } ∩ { 1 < x < 4 } ∩ { } = { 1 < x < 4 } Dengan demikian nilai x yang memenuhi adalah 1 < x < 4 . Oleh karena itu, jawaban yang benar adalah C.