How do you solve #lnx + ln (x-2) = 1#?

1 Answer
Gió
May 5, 2016

I found: #x=1+sqrt(1+e)#

Explanation:

I would write it as one log (multiplying the arguments) as:
#ln[x(x-2)]=1#
then use the definition of log and write:
#x(x-2)=e^1#
#x^2-2x-e=0#
use the Quadratic Formula to find #x#:
#x_(1,2)=(2+-sqrt(4+4e))/2=(2+-2sqrt(1+e))/2=1+-sqrt(1+e)#
I would use the positive one only, i.e.:
#x=1+sqrt(1+e)#

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