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anonymous
 one year ago
Hi everyone : hope ur well so I have this log plus absolute question so. For ln (x2) = ln (2x) I know there is no solution however for ln x2 = ln 2x There is , someone can kindly how is that possible and also help me solve this thank u
anonymous
 one year ago
Hi everyone : hope ur well so I have this log plus absolute question so. For ln (x2) = ln (2x) I know there is no solution however for ln x2 = ln 2x There is , someone can kindly how is that possible and also help me solve this thank u

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0how did you get *no solution* ???

alekos
 one year ago
Best ResponseYou've already chosen the best response.0i would imagine that there is a solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well when u do the inequality of x2 and 2x 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
 one year ago
Best ResponseYou've already chosen the best response.0However for absolute the answer is always positive

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm\cancel{ ln} (x2) = \cancel{\ln}(2x)\] x2 =2x isn't it right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope u have to check their inequality first then check if their 11 function then u can do the step u did

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.00_o okay? o_0 i should leave... ;P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 can u plz help me

alekos
 one year ago
Best ResponseYou've already chosen the best response.0If 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 \[\lim_{x \rightarrow 2} \ln \left x2 \right = \lim_{x \rightarrow 2} \ln \left 2x \right\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.00_o lim .. i'm sorry @ayeshaafzal221 __
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