How do you evaluate #3 log_2 (6) - 3/4 log_2 (81)#?

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How do you evaluate #3 log_2 (6) - 3/4 log_2 (81)#?

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Answer 1 · 2016-05-22T20:06:25

It is

#3 log_2 (6) - 3/4 log_2 (81)=3log_2 6-3/4*log_2 3^4= log_2 6^3-3/4*4*log_2 3=log_2 6^3-3/(cancel4) (cancel4)*log_2 3= log_2 6^3-log_2 3^3=log_2 (6/3)^3= log_2 (2)^3=3*log_2 2=3#

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How do you evaluate #3 log_2 (6) - 3/4 log_2 (81)#?

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