How do you condense #3log_2 b - log_2 4 - 2log_2 C#?

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Answer 1 · 2016-06-14T12:31:05

#3log_2b-log_2(4)-2log_2c=log_2(b^3/(4c^2))#

Using #plog_am=log_am^p# and #loga+logb-logc=log((ab)/c)#

#3log_2b-log_2(4)-2log_2c#

= #log-2(b^3)-log_2(4)-log_2(c^2)#

= #log_2(b^3/(4c^2))#

Answer 2 · 2016-06-14T12:43:00

#log_2(b^3/(4C^2)).#

The Rule (1) #mlog_2n=log_2n^m.#
Rule (2) #log_2p-log_2q=log_2(p/q).#

Thus, by Rule (1), #3log_2b=log_2b^3, -2log_2C=-log_2C^2.#
Combining these by Rule (2) we get, #3log_2b-2log_2C=log_2{(b^3)/(C^2)}.#

Finally, we get the simplification #=log_2{(b^3)/(C^2)}-log_2(4)=log_2{(b^3/C^2)/4}=log_2(b^3/(4C^2)).#