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How do you calculate #Log_sqrt(17)152#?

1 Answer

A. S. Adikesavan

Aug 3, 2016

#2 (log 152/log 17)= 3.5464#, nearly.

Explanation:

Use #log_b a=log a/log b#.

#log_sqrt 17 152#

#=log 157/logsqrt 17#

#=log 167/log(17^(1/2))

#=2 (log 152/log 17)=3.5464, #nearly.

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