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Math Help - ln and e

  1. #1
    Member Furyan's Avatar
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    ln and e

    The question is:

    Given that \ln(2x^2+9x-5)=1+\ln(x^2+2x-15), find x in terms of e

    I got as far as:

    e = \dfrac{2x-1}{x-3}

    The next step is

    e = \dfrac{2x-1}{x-3} \Rightarrow 3e - 1 = x(e - 2)

    I'm probably just being dense, but I don't know how to do that step. Could somebody please help me.

    Thank you
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  2. #2
    MHF Contributor
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    Re: ln and e

    x = \frac{3e - 1}{e - 2}

    and your work is correct.
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  3. #3
    Member Furyan's Avatar
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    Re: ln and e

    Quote Originally Posted by icemanfan View Post
    x = \frac{3e - 1}{e - 2}

    and your work is correct.
    Thanks for your reply. I see the last step, but I don't know how to do the step I've shown in original post.
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  4. #4
    Super Member Quacky's Avatar
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    Re: ln and e

    Quote Originally Posted by Furyan View Post
    The question is:

    Given that \ln(2x^2+9x-5)=1+\ln(x^2+2x-15), find x in terms of e

    I got as far as:

    e = \dfrac{2x-1}{x-3}

    The next step is

    e = \dfrac{2x-1}{x-3} \Rightarrow 3e - 1 = x(e - 2)

    I'm probably just being dense, but I don't know how to do that step. Could somebody please help me.

    Thank you
    e = \dfrac{2x-1}{x-3}

    Multiply through by (x-3):

    e(x-3)=2x-1

    Expand:

    ex-3e=2x-1

    Rearrange, so that terms involving an " x" are isolated on one side of the equation:

    ex-2x=3e-1

    Factor:

    x(e-2)=3e-1
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  5. #5
    Member Furyan's Avatar
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    Re: ln and e

    Quote Originally Posted by Quacky View Post
    e = \dfrac{2x-1}{x-3}

    Multiply through by (x-3):

    e(x-3)=2x-1

    ex-3e=2x-1

    Rearrange, so that terms involving an " x" are isolated on one side of the equation:

    ex-2x=3e-1

    Factor:

    x(e-2)=3e-1
    Thanks Quacky for taking the time to reply, that was what I was asking. Normally I'd see that, I assumed the rearrangment was some mysterious property of e. All these new numbers are scaring the sense out of me , although it was very late, I have so got to get a grip.

    Thanks again
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