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IF x=198! then value of the expression 1...

IF `x=198!` then value of the expression `1/(log_2x)+1/(log_3x)+...+1/(log_198 x)` equals

A

`-1`

B

0

C

1

D

198

Text Solution

Verified by Experts

The correct Answer is:
C
We have,
`(1)/("log"_(2)x) + (1)/("log"_(3)x) + (1)/("log"_(4)x) +…+ (1)/("log"_(198)x)`
` ="log"_(x) 2 + "log"_(x) 3 + "log"_(x) 4+…+ "log"_(x)198`
` = "log"_(x) 2 xx 3 xx 4 xx …. xx 198 = "log"_(x) 198! = "log"_(x) x =1`
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