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If `"log"_(a) x xx "log"_(5)a = "log"_(x) 5, a ne 1, a gt 0, " then "x =`A. aB. `5, (1)/(5)`C. 1D. none of these |
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Answer» Correct Answer - B We have, `"log"_(a) x xx "log"_(5)a = "log"_(x)5` `rArr "log"_(5) x= "log"_(x) 5` `rArr "log"_(5)x = (1)/("log"_(5)x)` `rArr ("log"_(5)x)^(2) = 1 rArr "log"_(5)x = +-1 rArr x = 5^(+-1) rArr x = 5, (1)/(5)` |
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