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Thread: Pre-Calc Help

  1. #1
    Ultimate Member shawshank62's Avatar
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    Pre-Calc Help

    Well i need to do this online assesment of pre-calc question, there were about 60 questions total, and the only ones i have left are the following three. I hoping someone here can point me in the right direction because im not really sure how to do them.

    Algebraic Properties of the Logarithm - Simplifying Formulas
    4) Simplify ln(z^4 + z^2).
    For this one i was thinking:
    4 ln(z) + 2 ln(z) but it says this answer is incorrect

    Algebraic Properties of the Logarithm - simplifying numbers
    1) Simplify ln (3000).
    I know that to do this i need to seperate it into ln( z^x + q^t), but i cant for the life of me figure out what number add up properly to 3000.

    Exponential Equations
    12) Solve the following equation in x: 2 e^(x) = e^(4x - 2)
    For this one i thought that i could take the 2 and move it to the x, so it would be

    e^(x)^(2) = e^(4x-2) and then multiply each side by ln(x) to get

    x^2 = 4x - 2 ------> x^2 - 4x - 2 but this doesnt appear to be the right answer so i know i went wrong somewhere.


    Any help would be greatly appreciated : )

  2. #2
    Ultimate Member shawshank62's Avatar
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    someone has to know more math then i do! where are you math people?

  3. #3
    F@H shizzle. xelnanga's Avatar
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    sorry. i know up to math b....

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    Ultimate Member cryptoguy's Avatar
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    I'm in Pre-Calc now, but we did natural logs last year and I'll be damned if I remember them

    And actually, number 4 seems right to me...
    23. That's the number of people Mr. T has pitied in the time it has taken you to read this sentence.

  5. #5
    Human voltmeter DanU's Avatar
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    ln(a+b) does not equal ln(a) + ln(b) !!!
    you must be thinking of ln(ab) = ln(a) + ln(b)

    OK I'll take a stab....

    Simplify ln(z^4 + z^2)
    factor the polynomial inside the () : ln ( (z^2)(z^2+1) )
    split into separate ln terms : 2*ln (z) + ln (z^2+1)
    you can reduce further, but you need to use the imaginary number "i": 2*ln (z) + ln((z+i)(z-i))
    and again separate terms : 2*ln (z) + ln(z+i) + ln(z-i)

    Simplify ln (3000)
    hmm... I'm not sure exactly what's wanted here. Personally I find ln(3000) to be "simple" enough... but anyway let's see what can be done:
    factor 3000 : ln (3*10^3)
    more factoring: ln (3 * 5^3 * 2^3)
    split into separate ln terms : ln(3) + 3*ln(5) + 3*ln(2)
    not really sure if that's what you wanted though....

    2 e^(x) = e^(4x - 2)
    rewrite so you can easily take the ln of both sides: e(x^2)=e^(4x-2)
    now take the ln of both sides: x^2 = 4x-2
    rewrite into quadratic equation form: 0=x^2-4x+2
    this polynomial is not easy to factor so you have to use the quadratic formula:
    if 0 = ax^2 + bx + c) then
    x = [-b +- sqrt(b^2 - 4ac)] / [2a]
    pluggin in a, b, and c you get
    x=[4 +- sqrt(16-4(1)2)]/[2(1)]
    x=[4 +- sqrt(8)]/2
    x=[4 +- 2sqrt(2)]/2

    x1= 2 + sqrt(2)
    x2= 2 - sqrt(2)

    hopefully I didn't screw up somewhere...
    Last edited by DanU; January 30th, 2006 at 12:44 AM.

  6. #6
    Ultimate Member shawshank62's Avatar
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    Thank you so much! i am ever in your debt lol

    Everything worked out perfect except for the last solution, but after doing it out i got the same exact answer as you so im not really sure whats going on with that problem. But only missing one problem out of all them isnt bad at all. So thanks again, you rock

  7. #7
    Human voltmeter DanU's Avatar
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    No problemo. Hope you get an "A"

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