How do you evaluate #log_3 (1/81)#?

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Answer 1 · 2016-10-18T17:32:05

By reducing the logarithm using the laws of logs, and reducing the number to a power of the base the result follows. #=-4#

#color(red)(log_"3"(1/81)=log_"3"(1)-log_"3"81)#

since

#log_a(X/Y)=log_a(X)-log_a(Y)#

#color(red)(log_"3"(1/81)=0-log_"3"3^4)#

since #log_a1=0#

#AAa inRR#

#color(red)(log_"3"(1/81)=-4log_"3"3)#

since #log_aX^n=nlog_aX#

#color(red)(log_"3"(1/81)=-4)#

since #log_aa=1#