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

1 Answer
Gió
Feb 4, 2017

I found: #x=(2+sqrt(4+4e))/2=1+sqrt(1+e)#

Explanation:

We can change the sum of logs into a log of a product and write:
#ln[x(x-2)]=1#
use the definition of log:
#x(x-2)=e^1#
rearrange and solve for #x#:
#x^2-2x-e=0#
Use the Quadratic Formula:
#x_(1,2)=(2+-sqrt(4+4e))/2#
#x_1=(2+sqrt(4+4e))/2>0# YES
#x_2=(2-sqrt(4+4e))/2<0# NO

Related questions
See all questions in Logarithmic Models
Impact of this question
1715 views around the world
You can reuse this answer
Creative Commons License