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anonymous

  • one year ago

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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  1. Nnesha
    • one year ago
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    how did you get *no solution* ???

  2. anonymous
    • one year ago
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    For the first one

  3. anonymous
    • one year ago
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    .

  4. alekos
    • one year ago
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    i would imagine that there is a solution

  5. anonymous
    • one year ago
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    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

  6. anonymous
    • one year ago
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    However for absolute the answer is always positive

  7. Nnesha
    • one year ago
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    \[\huge\rm\cancel{ ln} (x-2) = \cancel{\ln}(2-x)\] x-2 =2-x isn't it right ?

  8. anonymous
    • one year ago
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    Nope u have to check their inequality first then check if their 1-1 function then u can do the step u did

  9. Nnesha
    • one year ago
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    0_o okay? o_0 i should leave... ;P

  10. anonymous
    • one year ago
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    @ganeshie8 can u plz help me

  11. anonymous
    • one year ago
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    @Abhisar

  12. alekos
    • one year ago
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    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 \[\lim_{x \rightarrow 2} \ln \left| x-2 \right| = \lim_{x \rightarrow 2} \ln \left| 2-x \right|\]

  13. Nnesha
    • one year ago
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    0_o lim .-. i'm sorry @ayeshaafzal221 _-_

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