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    Name:  yhhhh.png
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Size:  37.5 KB Hi, I get that the second equation can be simplified from the first equation since e^ln cancels out right? But there are two different curves although the equations are the same?

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    (Original post by coconut64)
    Name:  yhhhh.png
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Size:  37.5 KB Hi, I get that the second equation can be simplified from the first equation since e^ln cancels out right? But there are two different curves although the equations are the same?

    Thanks
    As x is put into Ln first in the first equation, the curve isn't defined for In 0 so it's asymptotic to x=2

    The values e is put to may also be rounded in the first equation, and this could have a large effect due to the rapid divergence of the exponential function.
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    (Original post by coconut64)
    Name:  yhhhh.png
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Size:  37.5 KB Hi, I get that the second equation can be simplified from the first equation since e^ln cancels out right? But there are two different curves although the equations are the same?

    Thanks
    It's treating the red one as  e^{(\ln (2-x))^2}+2 .
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    (Original post by coconut64)
    Hi, I get that the second equation can be simplified from the first equation since e^ln cancels out right? But there are two different curves although the equations are the same?

    Thanks
    As said by B_9710, you entered it wrong and it's being treated at \ln^2 (2-x)

    If you enter \ln((2-x)^2) then you will get the same curve. But if you put in 2\ln(2-x) you will get half a curve, I hope you can see why.
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    (Original post by B_9710)
    It's treating the red one as  e^{(\ln (2-x))^2}+2 .
    Not going to lie, I thought computers were smart enough to be able to deal with brackets properly nowadays.
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    (Original post by RDKGames)
    As said by B_9710, you entered it wrong and it's being treated at \ln^2 (2-x)

    If you enter \ln((2-x)^2) then you will get the same curve. But if you put in 2\ln(2-x) you will get half a curve, I hope you can see why.
    I still don't understand because you can simplifty e^2ln(x-2) to e^ln(x-2)^2 right?

    Thanks
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    (Original post by coconut64)
    I still don't understand because you can simplifty e^2ln(x-2) to e^ln(x-2)^2 right?

    Thanks
    Let's pick an easier example. Let y_1=\ln(x^2) and y_2=2\ln(x)

    Now these are technically not the same functions depending on what domain you specify. If y_1 is defined on (0,+\infty) then y_1 \equiv y_2 but if for both functions they are given the domain of \{ \mathbb{R} ; x\not= 0 \} then y_1 is only defined for x>0 since \ln(x) doesnt take in negative x values, while \ln(x^2) does as they are squared anyway so they become positive.

    Now, the only way that y_1 \equiv y_2 across all reals (except 0) is if you define y_1=\ln \lvert x^2 \lvert and y_2=2\ln \lvert x \lvert

    (Original post by h3rmit)
    Not going to lie, I thought computers were smart enough to be able to deal with brackets properly nowadays.
    It's a computer - you need to tell it EXACTLY what it needs to do especially concerning mathematical notation and its possible ambiguity. Believe it or not, but on mathematical software such as Maple and MATLAB, instead of ab you need to write a*b
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    Thank you all for the help
 
 
 
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