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Math Help - E problem

  1. #1
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    Apr 2010
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    E problem

    Must find x

    Ae^Bx = e^Cx

    Where A,B,C are constants

    My work:

    lnA + Bx = Cx

    lnA / (C - B) = x

    I am not confident in this answer
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    I get

    Ae^{Bx} = e^{Cx}

    \ln (Ae^{Bx}) = \ln (e^{Cx})

    \ln (A) +\ln (e^{Bx})= \ln (e^{Cx})

    \ln (A) +Bx= Cx

    \ln (A) = Cx-Bx

    \ln (A) = x(C-B)

    \frac{\ln (A)}{C-B} = x

    x=\frac{\ln (A)}{C-B}

    Same as you, with are either both right or both wrong!
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  3. #3
    Super Member Bacterius's Avatar
    Joined
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    Wellington
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    Let me try

    Ae^{Bx} = e^{Cx}

    \ln{(Ae^{Bx})} = \ln{(e^{Cx})}

    \ln{(A)} + \ln{(e^{Bx})} = \ln{(e^{Cx})}

    \ln{(A)} + Bx = Cx

    \ln{(A)} = Cx - Bx

    \ln{(A)} = x(C - B)

    \frac{\ln{(A)}}{C - B} = x

    x = \frac{\ln{(A)}}{C - B}

    Wolfram says so too. Looks like this is the answer

    EDIT : Strange, I did exactly the same steps as Pickslides and in the same order ... but I didn't look at all.
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  4. #4
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    Thank you, pickslides and Bacterius, I appreciate the confirmation
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