Hi everyone : hope ur well so I have this log plus absolute question so. For ln (x-2) = ln (2-x) I know there is no solution however for ln |x-2| = ln |2-x| There is , someone can kindly how is that possible and also help me solve this thank u

Question

Hi everyone : hope ur well so I have this log plus absolute question so. For ln (x-2) = ln (2-x) I know there is no solution however for  ln |x-2| = ln |2-x| There is , someone can kindly how is that possible and also help me solve this thank u

nnesha · Answer

how did you get *no solution* ???

anonymous · Answer

For the first one

anonymous · Answer

.

alekos · Answer

i would imagine that there is a solution

anonymous · Answer

Well when u do the inequality of x-2 and 2-x it states that x is bigger than two but at the same time less than two but in reality that's not possible thus no solution

anonymous · Answer

However for absolute the answer is always positive

nnesha · Answer

$$\huge\rm\cancel{ ln} (x-2) = \cancel{\ln}(2-x)$$
x-2 =2-x 
isn't it right ?

anonymous · Answer

Nope u have to check their inequality first then check if their 1-1 function then u can do the step u did

nnesha · Answer

0_o okay? o_0
i should leave... ;P

anonymous · Answer

@ganeshie8  can u plz help me

anonymous · Answer

@Abhisar

alekos · Answer

If you plot a graph of both functions you will see that as x approaches 2 then y tends to +inf so there really is no valid solution except to say that 

$$\lim_{x \rightarrow 2} \ln \left| x-2 \right| = \lim_{x \rightarrow 2} \ln \left| 2-x \right|$$

nnesha · Answer

0_o lim .-. i'm sorry @ayeshaafzal221
_-_