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How do you evaluate #log_169 13#?

1 Answer

A. S. Adikesavan

Aug 10, 2016

#1/2#

Explanation:

Use log_b a=log_c a/log_c#

Here,
# log_169 13#

#=ln 13/ln 169#

#=ln 13/ln 13^2#

#=ln 13/(2ln 13)#

#=1/2#.

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