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Math Help - Diff. eqn almost solved. Can't get to next step using ln rules...

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    Diff. eqn almost solved. Can't get to next step using ln rules...

    I've been working through a midterm practice problem and I ran into a little hitch with one. The answer it gives me is the last few steps of the equation but I am stuck between two of them.

    The 2nd last step is a rearrange starting with -1/3ln|15-3I|=t+C I got to this step without an issue

    Getting from the 2nd last step to the last is where it confuses me. I(t)=5-Cexp(-3t)

    I multiplied the right side by -3 to give ln|15-3I|=3t+3C
    then exponentiated to 15-3I=exp(3t)+exp(3C)
    then -3I=exp(3t)+exp(3C)-15
    and I=(-1/3)exp(3t)+exp(3C)-5

    I'll admit my ln skills are getting a little rusty and maybe I've missed something obvious but I'm seriously stuck. Any help would be greatly appreciated. Thanks!
    -Q
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by QNev View Post
    I've been working through a midterm practice problem and I ran into a little hitch with one. The answer it gives me is the last few steps of the equation but I am stuck between two of them.

    The 2nd last step is a rearrange starting with -1/3ln|15-3I|=t+C I got to this step without an issue

    Getting from the 2nd last step to the last is where it confuses me. I(t)=5-Cexp(-3t)

    I multiplied the right side by -3 to give ln|15-3I|=3t+3C
    shouldn't you have -3t - 3C on the right?

    then exponentiated to 15-3I=exp(3t)+exp(3C)
    here's where you really messed up

    \displaystyle \exp(-3t + 3C) \ne \exp (-3t) + \exp (3C)

    it is in the power, you get this from MULTIPLYING like bases.

    \displaystyle \exp (-3t + 3C) = \exp (-3t) \cdot \exp (3C)

    now continue with that
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    Woops... Definitely typed that in wrong. I had it right on paper.

    15-3I=exp(-3t)*exp(-3C)
    I=(-1/3)exp(-3t)*exp(-3C)+5

    Not sure how this works out to I=5-Cexp(-3t) (as the answer says)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by QNev View Post
    Woops... Definitely typed that in wrong. I had it right on paper.

    15-3I=exp(-3t)*exp(-3C)
    I=(-1/3)exp(-3t)*exp(-3C)+5

    Not sure how this works out to I=5-Cexp(-3t) (as the answer says)
    (-1/3)exp(-3C) is a constant, they simply wrote it as -C. Not sure why the minus, but it makes no difference.

    so they have \displaystyle I(t) = 5 - \underbrace{\frac 13e^{-3C} }_{\text{call this } C}e^{-3t}
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    Thanks a lot!
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