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Math Help - Integral, log and Inverse

  1. #1
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    Exclamation Integral, log and Inverse

    1. \int_1^2 \frac{x-4}{x^2} dx *ANSWERED*

    2. log_a(2^a) = \frac{a}{4}, then a = * ANSWERED*

    3.Inverse of y=2x^3+3x^2+7x+4 *ANSWERED*
    (i know how to do inverse, but this polynomial is throwing me off...)


    Last edited by yamuda; April 20th 2008 at 05:37 PM.
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  2. #2
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    Quote Originally Posted by yamuda View Post
    1. \int_0^2 \frac{x-4}{x^2} dx

    2. log_a(2^a) = \frac{a}{4}, then a =

    3.Inverse of y=2x^3+3x^2+7x+4
    (i know how to do inverse, but this polynomial is throwing me off...)

    Here is number 2

    rewrite the left hand side using the change of base formula

    \log_a2^a=\frac{\log_22^a}{\log_2a}

    Now we get

    \frac{\log_22^a}{\log_2a}=\frac{a}{4} \iff \frac{a}{\log_2a}=\frac{a}{4}

    4=\log_2a \iff 2^4=a

    Good luck.
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  3. #3
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    Quote Originally Posted by yamuda View Post
    1. \int_0^2 \frac{x-4}{x^2} dx *NO ANSWER YET*

    2. log_a(2^a) = \frac{a}{4}, then a = * ANSWERED*

    3.Inverse of y=2x^3+3x^2+7x+4 *NO ANSWER YET*
    (i know how to do inverse, but this polynomial is throwing me off...)

    for the first use termwise division

    \int_{0}^{2}\frac{x-4}{x^2}dx=\lim_{a \to 0} \left(\int_{a}^{2}\frac{1}{x}dx-\int_a^2\frac{4}{x^2}dx \right)

    after integrating and evaluating the limit you will find that it divereges to infinity.
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  4. #4
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    hmmm...

    For the last one are supposed to find the inverse of just show that an inverses exisis?

    Here is a link to the general cubic formula( like the quadratic, but for degree 3 equations)

    The Cubic Formula

    Good luck
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    for the first use termwise division

    \int_{0}^{2}\frac{x-4}{x^2}dx=\lim_{a \to 0} \left(\int_{a}^{2}\frac{1}{x}dx-\int_a^2\frac{4}{x^2}dx \right)

    after integrating and evaluating the limit you will find that it divereges to infinity.
    mmm... i messed up on the integral... was supposed to be \int_1^2, im tryna figure out the integral, coming out with a numerical answer, not if it goes to infinity..

    and for the last one, im trying to find the inverse, so i switched y and x, then i have to solve for y
    this is all i got... x=2y^3+3y^2+7y+4
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  6. #6
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    Quote Originally Posted by yamuda View Post
    mmm... i messed up on the integral... was supposed to be \int_1^2, im tryna figure out the integral, coming out with a numerical answer, not if it goes to infinity..

    and for the last one, im trying to find the inverse, so i switched y and x, then i have to solve for y
    this is all i got... x=2y^3+3y^2+7y+4
    \int_{1}^{2}\frac{x-4}{x^2}dx=\int_{1}^{2}\frac{1}{x}dx-\int_1^2\frac{4}{x^2}dx=\ln|x|+\frac{4}{x}|_1^2

     <br />
\ln2+\frac{4}{2}- \left( \ln1+\frac{4}{1}\right)=\ln2-2<br />

    as for your last question the only way I know of is the formula from the above link.

    Good luck
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    \int_{1}^{2}\frac{x-4}{x^2}dx=\int_{1}^{2}\frac{1}{x}dx-\int_1^2\frac{4}{x^2}dx=\ln|x|+\frac{4}{x}|_1^2

     <br />
\ln2+\frac{4}{2}- \left( \ln1+\frac{4}{1}\right)=\ln2-2<br />

    as for your last question the only way I know of is the formula from the above link.

    Good luck
    alright, thanks for all the help!!!!
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