How do you solve #ln 3x+ ln 2x=3#?

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Answer 1 · 2015-12-14T08:56:26

#x=sqrt(frac{e^3}{6})#

Use the fact that #ln(ab)=ln(a)+ln(b)# for positive #a# and #b#.

#ln3x+ln2x=(ln3+lnx)+(ln2+lnx)#

#=ln6+2lnx#

#=3#

Perform algebraic manipulation yields

#lnx=(3-ln6)/2#

#x=e^(frac{3-ln6}{2})#

#=frac{e^(frac{3}{2})}{e^(frac{ln6}{2})}#

#=sqrt(frac{e^3}{6})#