A) \[{{2}^{\frac{1}{31}}}\]
B) \[{{2}^{\frac{1}{32}}}\]
C) \[{{2}^{\frac{31}{32}}}\]
D) \[{{2}^{\frac{30}{31}}}\]
Correct Answer: C
Solution :
(b): \[y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}\] \[{{y}^{2}}=2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}\] \[{{y}^{4}}=8\sqrt{2\sqrt{2\sqrt{2}}}\] \[{{y}^{8}}=128\sqrt{2\sqrt{2}}\] \[{{y}^{16}}={{\left( 128 \right)}^{2}}\times 2\sqrt{2}\] Then \[{{y}^{32}}={{\left( 128 \right)}^{4}}\times 4\times 2\] \[{{y}^{32}}={{2}^{28+2+1}}={{2}^{31}}\] \[{{2}^{\frac{31}{32}}}\]
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