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How do you solve for x in ##1/3lnx+ln2-ln3=3##?

The answer is: ##x=27/8e^9##.

First of all we have to add a condition otherwise our equation loses meaning: ##x>0##.

Than:

##1/3lnx+ln2-ln3=3rArrlnx=3(ln3-ln2+3)rArr##

##lnx=3(ln3-ln2+lne^3)rArrlnx=3ln(3e^3/2)rArr##

##lnx=ln(27/8e^9)rArrx=27/8e^9##

(That is positive, so it is acceptable).

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