How do you solve #ln x + ln (x-3) =0#?

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Answer 1 · 2016-07-29T20:51:23

#x = (3 + sqrt(13))/2#

#log_e x+log_e(x-3) = log_e 1# so
#log_e x(x-3) = log_e 1#

or

#x(x-3)=1 -> x^2-3x-1=0#

with the solutions

#x = (3 pm sqrt(13))/2# now, cheking the feasibility we choose

#x = (3 + sqrt(13))/2# because with this value,

#x > 0# and #x-3 > 0#