How do you calculate #Log_5 16#?

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Answer 1 · 2018-05-17T05:02:47

#color(maroon)(=> 1.7227#

#log_5 16#

#log_b a = log_x a/ log _x b#, applying change of base rule.

Let #b = 10#

#=> log 16 / log 5#

#=> log 2^4 / log 5#

#=> (4 log 2 )/ log 5" "larr# use a calculator or tables from here

#=> color(maroon)(1.7227#