ln(x+1)-lnx=e

I have to solve for x.  The problem I'm having is the - throws me off.
 This is what I've come up with.

I distribute the lnto the (x+1) to come up with
lnx+ln1-lnx=e
the lnx's cancel eachother out and therefore I cannot solve for x.

I'm looking for some instructions as to how to solve this.

Hi xkyroutx-ga,

Natural logarithm (ln) can not be distributed like what you have done:
you can not separate ln(x+1) into ln(x)+ln(1).

The following are some logarithmic identities that you should know:
1. log(A*B) = log(A)+log(B)
2. log(A/B) = log(A)-log(B)
3. log(A^B) = B * log(A)

Now, let's look at your equation:
ln(x+1)-ln(x) = e
By identity #2 above, you can simplify ln(x+1)-ln(x) into ln((x+1)/x), giving:
ln((x+1)/x) = e
The next is to take the exponent of e on both sides:

e^(ln((x+1/x) = e^e
(x+1)/x = e^e, x not equal to 0

now we multiply both sides by x, since x is not 0:
x+1 = x * e^e
and subtract x from both sides:
1 = x * e^e - x
1 = x(e^e - 1)
and divide both sides by e^e - 1:

x = 1/(e^e - 1)

Now to check the answer:

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

(x+1)/x = (1/(e^e-1) + 1)*(e^e-1)
(x+1)/x = 1 + e^e - 1 = e^e
ln ((x+1)/x) = ln (e^e) = e

So your answer is x = 1/(e^e - 1).

Reference:
Logarithmic identities: 
http://en.wikipedia.org/wiki/Logarithm#Easier_computations

I hope that this was clear.  If you need clarification, please use the
request for clarification feature before rating this answer.

secret901-ga