How do you evaluate #log_3 18#?

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Answer 1 · 2016-05-19T01:44:00

#= log_3(9 xx 2)#

#= log_3(9) + log_3(2)#

#= log9/log3 + log2/log3#

#= log3^2/log3 + log2/log3#

#= 2cancel(log3)/cancel(log3) + log2/log3#

#= 2 + log_3(2)#

Hopefully this helps!