Which are solutions of equation lnx + ln(x+1)=2?

Expert Answers
hala718 eNotes educator| Certified Educator

lnx+ln(x+1)=2

We know that lnxy= lnx+lny

==> lnx+ln(x+1) = lnx(x+1)=2

==> ln(x^2+x) = 2

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

==> x^2+x-e^2=0

==> x1=[-1+(1+4e^2)^1/2]/2

       x2=[-1-(1+4e^2)^1/2]/2

However, we will not consider x2 since it is a negative value.

Then the solution is x1= [-1+(1+4e^2)^1/2]/2

giorgiana1976 | Student

We'll use the product 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 opening the brackets, we'll get:

x^2 + x - e^2 = 0

We'll calculate the roots applying the quadratic formula:

x1=[-1+sqrt(1+4e^2)]/2

x2=[-1-sqrt(1+4e^2)]/2

Before stating which are the solutions of the equation, let's recall the constraints of a logarithmical function.

x>0 and x+1>0

So, x>0

It is obvious that x2 is not a valid solution because it's negative.

The only solution of the equation is:

x1=[-1+sqrt(1+4e^2)]/2

neela | Student

To solve for lnx+ln(x+2)  = 2.

Solution:

lnx+ln(x+2) = ln[x(x+2)] is given to be equal to 2.

Or

log[x(x+2)] = 2 = log e^2. So x(x+2) -e^2 = 0. Or

x^2+x-e^2 = 0.This is a quadratic equation

x1 = {-1+sqrt(1+4e^2)}/2 Or

x2 = {-1-sqrt(1+4e^2)}/2