How do you solve #ln x = 1 - ln (x+8)#?

2 Answers
Aug 6, 2015

I found: #x=-4+sqrt(16+e)#

Explanation:

Write it rearranging as:
#ln(x)+ln(x+8)=1#
use the fact that #lnx+lny=ln(xy)#
so:
#ln[x(x+8)]=1#
use the definition of log:
#lnx=a -> x=e^a#
#x(x+8)=e^1#
#x^2+8x-e=0#
using the Quadratic Formula:
#x_(1,2)=(-8+-sqrt(64+4e))/2=(-8+-2sqrt(16+e))/2=#
#=-4+-sqrt(16+e)#
so you get two solutions but one, #-4-sqrt(16+e)#, doesn't work when substituted into the original equation (it is a negative number, try it!) and you can discard it keeping only:
#x=-4+sqrt(16+e)#

Aug 6, 2015

x=#-4+-sqrt(16+e)#

Explanation:

Rewriting the equation , it is ln x + ln(x+8) =1

ln x(x+8)=1

#x^2 +8x= e#

#x^2 +8x -e=0#

x= #(-8+-sqrt (64+4e))/2#

x=#-4+-sqrt(16+e)#