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

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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

Mathematics
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how did you get *no solution* ???
For the first one
.

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i would imagine that there is a solution
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
However for absolute the answer is always positive
\[\huge\rm\cancel{ ln} (x-2) = \cancel{\ln}(2-x)\] x-2 =2-x isn't it right ?
Nope u have to check their inequality first then check if their 1-1 function then u can do the step u did
0_o okay? o_0 i should leave... ;P
@ganeshie8 can u plz help me
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 \[\lim_{x \rightarrow 2} \ln \left| x-2 \right| = \lim_{x \rightarrow 2} \ln \left| 2-x \right|\]
0_o lim .-. i'm sorry @ayeshaafzal221 _-_

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