How do you solve #ln4x+lnx=9#?

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Answer 1 · 2016-11-16T18:15:52

Use the product rule for logarithms you get

#ln4x+lnx=9#

#ln4x^2=9#

#4x^2=e^9#

#(2x)^2=e^9#

#2x=sqrt(e^9)#

#x=sqrt(e^9)/2#

Remark : from the defination of the logarithm we have #x>0#