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If `9^("log"_3("log"_(2) x)) = "log"_(2)x - ("log"_(2)x)^(2) + 1,` then x =A. 1B. `1//2`C. 3D. none of these |
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Answer» Correct Answer - D We have, `9^("log"3("log"_(2) x)) = "log"_(2)x - ("log"_(2)x)^(2) + 1` `rArr 3^(2"log"_(3) ("log"_(2)x)) = "log"_(2) x - ("log"_(2)x)^(2) +1` `rArr 3^("log"_(3) ("log"_(2) x)^(2)) = "log"_(2) x - ("log"_(2)x)^(2) + 1` `rArr ("log"_(2) x)^(2) = "log"_(2)x - ("log"_(2)x)^(2) + 1` `rArr 2("log"_(2) x)^(2) - "log"_(2) x - 1 =0` `rArr (2"log"_(2) x +1) ("log"_(2)x-1) = 0` `rArr "log"_(2)x = -(1)/(2), "log"_(2) x =1 rArr x = 2^(-1//2), 2` |
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