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If `a_(n) gt a_(n-1) gt …. gt a_(2) gt a_(1) gt 1`, then the value of `"log"_(a_(1))"log"_(a_(2)) "log"_(a_(3))…."log"_(a_(n))` is equal to |
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Answer» Correct Answer - B We have, `"log"_(a_(1)) "log"_(a_(2)) "log"_(a_(3))…. "log"_(a_(n-1)) "log"_(a_(n)) a_(n)^(a_(n-1)^(" "a_(n-2)^(" ".^(.^(.^(a_(1)))))))` ` = "log"_(a_(1)) "log"_(a_(2)) "log"_(a_(3))…. "log"_(a_(n-1)) a_(n-1)^(" "a_(n-2)^(" ".^(.^(.^(a_(1))))))` `= "log"_(a_(1))"log"_(a_(2))"log"_(a_(3))..... "log"_(a_(n-2)) a_(n-2)^(" "a_(n-3)^(" ".^(.^(.^(a_(1))))))` `=...= "log"_(a_(1)) a_(1) =1` |
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