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))#