How do you condense $2logx-log4-log3+logx$ ?

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Answer 1 · 2016-09-01T23:23:00

$log (x^3/12)$

If logs are added, the numbers are multiplied.
If logs are subtracted, the numbers are divided.

$2logx-log4-log3+logx$

= $logx^2 + logx -log4-log3$

= $log[(x^2 xx x)/(4xx3)]$

= $log (x^3/12)$

How do you condense 2logx-log4-log3+logx2logx-log4-log3+logx?