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Math Help - does anyone know wht this means?

  1. #1
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    does anyone know wht this means?

    Hey, im doing data management for a engineering lab, and one of the questions asks to prove that doing the semi-log plot of a graph with exponential correlation give the same relation. A semi-log graph is taking all the y values of the exponential graph and applying the natural logirthm to it.
    ex: y= 6 in the semi long graph y= ln 6
    This will give a linear relationship which is easier to work with.
    My question is, they ask me to prove this, by taking the regression equation algebraically alter it to equal thre regression equation of the linear relation ship
    this isthe exponential relation:

    1E-230e^(0.2682x)
    What exactly does the E mean, and how can i manipulate this to equal,

    0.1165x 229.96
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lmao6 View Post
    Hey, im doing data management for a engineering lab, and one of the questions asks to prove that doing the semi-log plot of a graph with exponential correlation give the same relation. A semi-log graph is taking all the y values of the exponential graph and applying the natural logirthm to it.
    ex: y= 6 in the semi long graph y= ln 6
    This will give a linear relationship which is easier to work with.
    My question is, they ask me to prove this, by taking the regression equation algebraically alter it to equal thre regression equation of the linear relation ship
    this isthe exponential relation:

    1E-230e^(0.2682x)
    What exactly does the E mean, and how can i manipulate this to equal,

    0.1165x 229.96
    i'm pretty sure the E means 10^, that's what it means on my calculator, for instance.

    if i enter 1.2 \times 10^2, then 1.2 E2 shows up on my display screen
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  3. #3
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    thank you SO much, that was bugging me ther whole day.
    I can finally solve this question many thanks
    btw im new to these forums, but i think i will be putting alot more time in here.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lmao6 View Post
    thank you SO much, that was bugging me ther whole day.I can finally solve this question many thanks
    you're welcome

    btw im new to these forums, but i think i will be putting alot more time in here.
    ok. good
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  5. #5
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    heh back for a bit more help...
    so this is what i tried:
    log(10^(-230e^(0.2682x))
    =-230e^(0.2682x)(1)
    =ln(-230e^(0.2682x))
    =ln (-230) + ln e^(0.2682x)
    =ln (-230) + 0.2682x
    almost got it! but im not sure what im doing wrong :S
    0.1165x 229.96 is what im trying to get
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