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

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Answer 1 · 2016-12-17T02:41:22

$log (\frac{x^{3} y^{4}}{z^{2}})$ $log ((x^3y^4)/z^2)$

Condense $3 log x + 4 log y - 2 log z$ $3logx+4logy-2logz$

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

First, use the log rule $a log x = {log x}^{a}$ $alogx=logx^a$

${log x}^{3} + {log y}^{4} - {log z}^{2}$ $logx^3+logy^4-logz^2$

Next, use the log rules

$log a + log b = log (a b)$ $loga +logb=log(ab)$ and $log a - log b = log (\frac{a}{b})$ $loga-logb=log(a/b)$

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

$log (\frac{x^{3} y^{4}}{z^{2}})$ $log ((x^3y^4)/z^2)$

How do you condense 3 log + 4 log y - 2 log z3log+4logy−2logz3 log + 4 log y - 2 log z?