How do you evaluate #log_18(9) + log_18(2)#?

2 Answers
Jim G.
Oct 29, 2016

#1#

Explanation:

Using the #color(blue)"laws of logarithms"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(logx+logy=log(xy)" and " log_b b=1)color(white)(2/2)|)))#

#rArrlog_(18)9+log_(18)2=log_(18)18=1#

Aditya Banerjee.
Oct 29, 2016

The answer is #log_18 (9)+log_18 (2)##=1#.

Explanation:

#log_18 (9)+log_18 (2)#
#=log_18 (9*2)#
#=log_18 (18)#
#=log18/log18#
#=1#. (answer).

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