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 $=log_2{(b^3)/(C^2)}-log_2(4)=log_2{(b^3/C^2)/4}=log_2(b^3/(4C^2)).$

How do you condense 3log_2 b - log_2 4 - 2log_2 C3log_2 b - log_2 4 - 2log_2 C?