For the beginning, we'll use the quotient property of the logarithms:

lnx - ln(x+1) = ln [x/(x+1)]

Now, we'll have to use the one to one property, that means that:

ln [x/(x+1)] = 2 lne if and only if [x/(x+1)] = e^2

After cross multiplying, we'll get:

x = x*e^2 + e^2

We'll move the terms which are containing the unknown, to the left side:

x - x*e^2 = e^2

After factorizing, we'll get:

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

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

But 1-e^2<0, so x = e^2/(1-e^2)<0, which is impossible because x has to be positive!

So, the equation has no solutions.

ln is used for denoting natural logarithm that is logarithm to the base e.

To solve the equation ln x - ln (x+1) = 2, use the property of logarithm, log a - log b = log(a/b)

This gives:

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

`log_e (x/(x +1)) = 2`

Now log_b a = c gives a = b^c

The equation can be written as:

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

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

`1 + 1/x = 1/e^2`

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

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

The solution of the equation `ln x - ln (x+1) = 2` is `x = 1/(1/e^2 - 1)`

To solve: lnx-ln(x+2) = 2.

Solution:

LHS = ln {x/(x+2)} = 2. Taking anti logarithms on both sides,

x/(x+2) = e^2. Or

x = e^2(x+2) . Or

x-xe^2 = 2e^2. Or

x(1-e^2) = 2e^2. Or

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