How do you solve #lnx-ln3=2#?

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Answer 1 · 2016-02-18T08:31:49

#x=3*e^2#

from the given: #ln x-ln 3=2#

#ln (x/3)=2#

also this means

#ln_e (x/3)=2#

taking the exponential form

#e^2=x/3#

#3*e^2=3*x/3# multiplying both sides by 3

#3*e^2=cancel3*x/cancel3#

#3*e^2=x#

and

#color (red)(x=3*e^2)#

Check: at #x=3*e^2# using the original equation

#ln x-ln 3=2#

#ln 3*e^2-ln 3=2#

#ln ((3*e^2)/3)=2#

#ln (e^2)=2#

#2=2#