How do you condense $2log a + log b – 3log c – log d$ ?

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Answer 1 · 2016-03-24T15:17:15

$log((a^2b)/(c^3d))$

First, bring the multiplicative constants into the logarithmic expressions using the rule:

$nlogx=log(x^n)$

Thus, we obtain:

$=log(a^2)+logb-log(c^3)-logd$

We now use the rule that combines added logarithms through multiplication of their arguments:

$logm+logn=log(mn)$

Applying this to $log(a^2)$ and $logb$ , this yields;

$log(a^2b)-log(c^3)-logd$

Opposite to adding, subtracted logarithms can have their arguments divided, in the rule:

$logm-logn=log(m/n)$

This gives the final answer of

$=log((a^2b)/(c^3d))$

How do you condense 2log a + log b – 3log c – log d2log a + log b – 3log c – log d?