Question

How do you solve `ln (x-3) - ln (x+4)=ln(x-1) - ln(x+2)`?

Answer 1

I found: NO REAL `x`.

Explanation:

You can start by using the property of logs that tells us:
`loga-logb=log(a/b)`
and write:
`ln((x-3)/(x+4))=ln((x-1)/(x+2))`
if the two logs must be equal their arguments must be as well; so:
`(x-3)/(x+4)=(x-1)/(x+2)` rearranging and getting rid of the denominators:
`(x-3)(x+2)=(x-1)(x+4)`
`cancel(x^2)+2x-3x-6=cancel(x^2)+4x-x-4`
isolate `x`:
`-x-3x=6-4`
`-4x=2`
`x=-2/4=-1/2`
If you plug this solution back into the original equation, you get some negative arguments of logs which prompt us to exclude `x=-1/2`.