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Math Help - Making x the subject in an exponential equation

  1. #1
    Member smmmc's Avatar
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    Making x the subject in an exponential equation

    make x the subject
    y=5(1-e^(-x))

    first i expand it

    so
    y=5-5e^(-x)

    loge(y)=loge(5)+loge(e5x)
    loge(y)-loge(5)=5x

    x=loge(y)-loge(5)/5 ??
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  2. #2
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    Quote Originally Posted by smmmc View Post
    make x the subject
    y=5(1-e^(-x))

    first i expand it

    so
    y=5-5e^(-x)

    loge(y)=loge(5)+loge(e5x)
    loge(y)-loge(5)=5x

    x=loge(y)-loge(5)/5 ??
    y = 5\left(1 - e^{-x}\right)

    \frac{y}{5} = 1 - e^{-x}

    e^{-x} = 1 - \frac{y}{5}

    e^{-x} = \frac{5 - y}{5}

    -x = \ln{\frac{5 - y}{5}}

    x = -\ln{\frac{5 - y}{5}}

    (Anything from this step onwards is just to get the answer to whatever form you like)

    x = \ln{\left(\frac{5 - y}{5}\right)^{-1}}

    x = \ln{\frac{5}{5 - y}}

    x = \ln{5} - \ln{(5 - y)}.
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  3. #3
    Member smmmc's Avatar
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    the answer says loge(5/5-y) ?
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  4. #4
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    Dear smmmc,

    Of course, \ln\frac{5}{5-y}=\ln5-\ln(5-y) Therefore what Prove it had shown is the answer.
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  5. #5
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    Quote Originally Posted by smmmc View Post
    the answer says loge(5/5-y) ?
    Which is one of the forms of the answer given by Prove It. And you have been told in another post what ln means. So what's your problem here?
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  6. #6
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    Quote Originally Posted by smmmc View Post
    make x the subject
    y=5(1-e^(-x))

    first i expand it

    so
    y=5-5e^(-x)

    loge(y)=loge(5)+loge(e5x)
    This is wrong. log(a+ b) is not equal to log(a)+ log(b)

    As others have said, the simplest thing to do is write y- 5= -5e^{-x+} and take the logarithm of that.
    loge(y)-loge(5)=5x

    x=loge(y)-loge(5)/5 ??
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