Solve the logarithmic equation lnx - ln(x+1) = 2?

4 Answers | Add Yours

hala718's profile pic

Posted on

ln x -ln (x+1) = 2

since ln x-ln y = ln x/y

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

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

==> (x+1)e^2 = X

==> xe^2 + e^2 = x

==> xe^2 -x= -e^2

==> x(e^2-1) = - e^2

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

giorgiana1976's profile pic

Posted on

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.

tonys538's profile pic

Posted on

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)`

neela's profile pic

Posted on

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)

We’ve answered 320,089 questions. We can answer yours, too.

Ask a question