How do you solve #ln(x+1)-ln(x-2)=ln x#?

Question

Benjamin Achtenberg · Accepted Answer

#x = 3/2 +sqrt(13)/2#

Taking the exponent of both sides, we get:

#(x+1)/(x-2) = x#

Multiplying both sides by #(x-2)# we get:

#x+1 = x^2-2x#

Subtract #x+1# from both sides to get:

#0 = x^2-3x-1#

#color(white)(0) = (x-3/2)^2-9/4-1#

#color(white)(0) = (x-3/2)^2-(sqrt(13)/2)^2#

#color(white)(0) = (x-3/2-sqrt(13)/2)(x-3/2+sqrt(13)/2)#

So:

#x = 3/2 +-sqrt(13)/2#

One of these values #3/2+sqrt(13)/2# is positive and a valid solution of the original equation.

The other value #3/2-sqrt(13)/2# is negative so not a solution if #ln# is the real logarithm. In fact it is not even a solution if we consider the complex logarithm, for which:

#ln(3/2-sqrt(13)/2+1) - ln(3/2-sqrt(13)/2-2)#

#= ln(5/2-sqrt(13)/2) - ln(1/2+sqrt(13)/2)-ipi#

#!= ln(sqrt(13/2)-3/2)+ipi = ln(3/2-sqrt(13)/2)#

Theodore Amidon · Answer

#(3+sqrt(13))/2#

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

If:

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

Then:

#(x+1)/(x-2)=x =x^2-3x-1#

Solving: #x^2-3x-1#

#x = (3+sqrt(13))/2 and x= (3-sqrt(13))/2#

# (3-sqrt(13))/2# not valid as gives negative value in #ln(x-2)#

So solution is:

#(3+sqrt(13))/2#

Ariana Alvarez · Answer

See below.

In the quest for real solutions, supposing #x ne {0,2}# #log_e abs(x+1)-log_e abs(x-2) = log_e absx# or #log_e(abs(x+1)/(abs(x-2)absx))=0# or #abs(x+1)/(abs(x-2)absx)=1# whose solutions are included in the solutions for #(x+1)^2=(x-2)^2x^2# or #(x+1-(x-2)x)(x+1+(x-2)x)=0# or #(1+3x-x^2)(1-x+x^2)=0# and the real solutions are from #1+3x-x^2=0# with the only feasible solution #x = 1/2 (3 + sqrt[13])#

Chloe Alexander · Answer

undefined To solve the equation $ \ln(x+1) - \ln(x-2) = \ln(x) $, follow these steps:

1. Apply the properties of logarithms to combine the logarithmic terms.
2. Use the properties of logarithms to simplify the equation.
3. Solve for $ x $ by isolating it on one side of the equation.
4. Check for extraneous solutions.

Combining the logarithmic terms:

$ \ln(x+1) - \ln(x-2) = \ln(x) $  
$ \ln\left(\frac{x+1}{x-2}\right) = \ln(x) $

Simplify the equation:

$ \frac{x+1}{x-2} = x $

Now, solve for $ x $:

$ (x+1) = x(x-2) $  
$ x + 1 = x^2 - 2x $  
$ x^2 - 3x - 1 = 0 $

Apply the quadratic formula to solve for $ x $:

$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $  
$ x = \frac{3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)} $  
$ x = \frac{3 \pm \sqrt{9 + 4}}{2} $  
$ x = \frac{3 \pm \sqrt{13}}{2} $

Thus, the solutions for $ x $ are $ x = \frac{3 + \sqrt{13}}{2} $ and $ x = \frac{3 - \sqrt{13}}{2} $.

However, it's important to check for extraneous solutions by substituting each solution back into the original equation and ensuring it satisfies the domain of the natural logarithm function.