SOLUTION: I need help with a couple equation I need to solve concerning logarithms: 1) {{{ln(x+2)=lne^ln2-lnx}}} I got that x=0 in this problem...but I am really confuses though

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I need help with a couple equation I need to solve concerning logarithms: 1) {{{ln(x+2)=lne^ln2-lnx}}} I got that x=0 in this problem...but I am really confuses though      Log On


   

Question 133119: I need help with a couple equation I need to solve concerning logarithms:
1) ln%28x%2B2%29=lne%5Eln2-lnx

I got that x=0 in this problem...but I am really confuses though
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
ln%28x+%2B+2%29+=+ln%28e%5Eln%282%29%29+-+ln%28x%29
We need to break this down:
y+=+ln%28e%5Eln%282%29%29
e%5Ey+=+e%5Eln%282%29
y+=+ln%282%29
ln%28e%5Eln%282%29%29+=+ln%282%29
So:
ln%28x+%2B+2%29+=+ln%282%29+-+ln%28x%29
ln%28x+%2B+2%29+=+ln%282%2Fx%29
x+%2B+2+=+2%2Fx
x%5E2+%2B+2x+=+2
x%5E2+%2B+2x+%2B+1+=+2+%2B+1
%28x+%2B+1%29%5E2+=+3
x+%2B+1+=+0%2B-sqrt%283%29
x+=+-1%2B-sqrt%283%29
The negative will result with a natural log of a negative, so only the positive term works: x+=+-1%2Bsqrt%283%29