How do you condense #3 log + 4 log y - 2 log z#?

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Answer 1 · 2016-12-17T02:41:22

#log ((x^3y^4)/z^2)#

Condense #3logx+4logy-2logz#

Note: I assumed there was a typo in the question and added an #x#.

First, use the log rule #alogx=logx^a#

#logx^3+logy^4-logz^2#

Next, use the log rules

#loga +logb=log(ab)# and #loga-logb=log(a/b)#

There is a somewhat silly expression for this rule: in the land of logs, addition is multiplication and subtraction is division.

#log ((x^3y^4)/z^2)#