HELP: ln(x+1) - ln x= 2

I don't understand this logarithms stuff.. :( It's an online course, so it's even more complex. Can you please show me out to work this out? Then I can do the rest!

ln(x + 1) - ln(x) = 2

By logarithmic rules, this can be re-written:

ln((x + 1) / x) = 2

Now, using e, eliminate the ln function:

e^ln((x + 1) / x) = e^2

(x + 1) / x = e^2

Now, break up the numerator to get:

x/x + 1/x = e^2

Since x/x = 1, we have:

1 + 1/x= e^2

Subtract 1 from each side:

1/x = e^2 - 1

Now, take the reciprocal of each side to get the final answer:

x = 1 / (e^2 - 1)

ok 1st i explain what a logarithm is.....

log properties:
ln(A)-ln(B)=ln(A/B)
ln(A)+ln(B)=ln(AB)
ln(A^B)=Bln(A)...this is A to the power B equals B times ln(A)
ln(e^A)=Aln(e) =A ....ln(e)=1 e is the base and the expponent that makes e=e is 1 so ln(e)=1
e^(ln(A))=A
ln(e^A)=A
now we can solve the problem using these properties

ln(x+1)-ln(x)=2
ln(x+1/x)=2
e^( ln(x+1/x) )=e^2 ....taking the exponential on both sides
x+1/x =e^2

1 +1/x =e^2

1/x =e^2 -1
therefore x = 1/(e^2 -1)

ln(x+1) - ln x= 2
ln[(x+1)/ x]= 2
( x+1)/x=e^2
x+1=xe²
xe²-x=1
x(e²-1)=1
x=1/(e²-1)
x=0.15652

ln(x + 1) - ln(x) = 2

ln(1 + 1/x) = 2

e^2 = 1 + 1/x

x = 1/(e^2 - 1)

x = 1/(-1 + e^2)