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

1 Answer
Jul 29, 2016

Answer:

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

Explanation:

#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#