How do you solve #ln x + ln (x+3) = 1#?

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Answer 1 · 2016-09-25T03:57:54

#x~=0.72896#

#lnx+ln(x+3)=1#

#ln(x*(x+3))=1#

#x(x+3) = e^1 = e#

#x^2+3x-e=0#

#x=(-3+-sqrt(9+4e))/2#

#~= (-3+-sqrt(9+10.87313))/2#

#~=(-3+-4.45793)/2#

Since #lnx# is defined for #x>0#

#x~= (-3+4.45793)/2#

#x~=0.72896#