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Thread: ln and e

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

    The question is:

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

    I got as far as:

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

    The next step is

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

    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

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

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

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

    I got as far as:

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

    The next step is

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

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

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

    Multiply through by $\displaystyle (x-3)$:

    $\displaystyle e(x-3)=2x-1$

    Expand:

    $\displaystyle ex-3e=2x-1$

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

    $\displaystyle ex-2x=3e-1$

    Factor:

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

    Multiply through by $\displaystyle (x-3)$:

    $\displaystyle e(x-3)=2x-1$

    $\displaystyle ex-3e=2x-1$

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

    $\displaystyle ex-2x=3e-1$

    Factor:

    $\displaystyle x(e-2)=3e-1$
    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 $\displaystyle e$. 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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