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The value of `x^("log"_(x) a xx "log"_(a)y xx "log"_(y) z)` isA. xB. yC. zD. a prime number |
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Answer» Correct Answer - C We have, `"log"_(x)a xx "log"_(a)y xx "log"_(y)z = "log"_(x)y xx "log"_(y)z = "log"_(x)z` `therefore x^("log"_(x)a xx "log"_(a)y xx "log"_(y)z) = x^("log"_(x)z) = z " " [because a^("log"_(a) N)=N]` |
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