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Precalc Help?


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#1 Random

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Posted 17 November 2008 - 06:23 PM

So I have a test tomorrow and I realize I have no idea how to do any of this garbage.

For example:
Solve for x, leave in ln or e if applicable:
ln x + ln(x^2 - 9) = ln 9x

or this one

Same directions:
e^2x + e^x - 12 =0

Can anyone help me with these? It would be much appreciated.

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    #2 Frank274

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    Posted 17 November 2008 - 07:47 PM

    I'm in precalc, and this doesn't make any sense to me.

    I'm doing polynomial dividing tongue.gif


    #3 rockage

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    Posted 17 November 2008 - 07:53 PM

    We did this in 3-4 last year, evaluating logs and all.

    Since it's a natural log (or at least from what I see they are) you know they are base 10, so that should help you out...


    #4 hungryhippo

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    Posted 17 November 2008 - 08:03 PM

    QUOTE (rockage @ Nov 17 2008, 10:53 PM) <{POST_SNAPBACK}>
    We did this in 3-4 last year, evaluating logs and all.

    Since it's a natural log (or at least from what I see they are) you know they are base 10, so that should help you out...


    that is not right. Natural logs are logs with base e

    #5 rockage

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    Posted 17 November 2008 - 08:08 PM

    Ah I meant common log, my mistake.

    #6 Melchoire

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    Posted 17 November 2008 - 08:18 PM

    I think I got the 2nd one:

    e2x + ex - 12 = 0
    e2x + ex = 12

    Take the natural log from both sides:
    2xlne + xlne = 12

    The natural long of e is 1 so:
    2x(1) + x(1) = 12
    3x = 12
    x = 4

    Edited by FlashGM, 17 November 2008 - 08:18 PM.


    #7 Casilla

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    Posted 18 November 2008 - 12:43 AM

    The great things about logs is that they have some simple expansion rules!

    log(xy) = logx + logy
    log(x/y) = logx - logy


    ln x + ln(x^2 - 9) = ln 9x

    ln(x) + ln(x^2 - 9) = ln(9) + ln(x)

    ln(x) cancels out from both sides

    ln(x^2 - 9) = ln (9)
    ln(x^2 - 9) - ln(9) = 0
    ln((x^2 -9)/9) = 0

    Now, we know ln(1) = 0, so therefore,

    (x^2 - 9)/9 = 1, which means x^2 - 9 = 9, x^2 = 18, so x has to be radical 18 (that is, square root of 18).

    And FlashGM got the second one.





    #8 ShadowLink64

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    Posted 18 November 2008 - 03:43 PM

    Oh, people beat me to this topic. sad.gif

    EDIT: I agree with Sonya's answer, but I wish to contest FlashGM's.

    e2x+ex - 12 = 0

    Write it in the form of a quadratic equation:
    (ex)2+ex - 12 = 0

    Factor it:
    (ex + 4)(ex - 3) = 0

    Therefore:
    e^x = -4
    and
    e^x = 3

    Take the natural logarithm of both sides (the first root is impermissible since ln(-something) does not exist in the domain of ln(x)):

    x = ln(-4) << Does not Exist
    x = ln(3)

    There is your solution... x = ln(3). Try it. tongue.gif

    QUOTE (FlashGM @ Nov 17 2008, 11:18 PM) <{POST_SNAPBACK}>
    Take the natural log from both sides:
    2xlne + xlne = 12


    ln(e^2x + e^x) cannot be divided into ln(e^2x) + ln(e^x). tongue.gif Only if there's a product inside your logarithm. wink.gif

    #9 Zip

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    Posted 18 November 2008 - 05:49 PM

    LOL nub.

    Keep in mind that e^ln(x) = x
    and
    ln(e^x) = x

    and it's rather simple.

    #10 Random

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    Posted 18 November 2008 - 06:41 PM

    Thanks guys biggrin.gif I think I did alright on my test even though I admittedly didn't see thse last night tongue.gif


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