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

1 Answer
Gió
Jul 28, 2016

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

Explanation:

We can change the sum of logs into a product of arguments as:
#ln[x(x-2)]=1#
We then use the definition of log to get:
#x(x-2)=e^1#
Rearranging:
#x^2-2x-e=0#
Use the Quadratic Formula:
#x_(1,2)=(2+-sqrt(4+4e))/2=(2+-sqrt(4(1+e)))/2=#
#=(2+-2sqrt(1+e))/2=(2(1+-sqrt(1+e)))/2=1+-sqrt(1+e)#
So:
#x_1=1+sqrt(1+e)#
#x_2=1-sqrt(1+e)# NO, because negative.

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