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If`y=a^(1/(1-(log)_a x))a n dz=a^(1/(1-(log)_a y)),t h e np rov et h a tx=a^(1/(1-(log)_a z))`A. `(1)/(a^(1-"log"az))`B. `(1)/(1-"log"_(a)z)`C. `(1)/(1+"log"_(z)a)`D. `(1)/(1-"log"_(z)a)` |
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Answer» Correct Answer - B We have, `y= a^((1)/(1-"log"_(a)z)) " and " z= a^((1)/(1-"log"_(a)y))` `rArr "log"_(a)y= (1)/(1-"log"_(a)x) " and log"_(a) z = (1)/(1-"log"_(a)y)` `rArr "log"_(a)x= ("log"_(a)y-1)/("log"_(a)y) " and "1-"log"_(a)z = ("log"_(a)y)/("log"_(a)y-1)` `rArr "log"_(a)x= (1)/(1-"log"_(a)z)` `rArr k = (1)/(1-"log"_(a)z) " " [because x = a^(k) therefore "log"_(a)x = k]` |
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