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Math Help - need help with some math questions

  1. #1
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    need help with some math questions

    1. x^1/3 = 32/ sqrt x 1/3 is the exponent

    2. if 9^x - 9^x-1 = 216, find 4^x x-1 is the exponent.

    3. find the inverse function of f(x) = 3sqrt (x + 1) ( on the left side of square root it has
    3 slightly on top so i guess it means whatever sqrt of x + 1 is it's to the power of 3.

    4. rationalize the denominator

    5x/sqrt (x) - sqrt (2x-5)

    5. prove that x-s-t is a factor of x^3 - s^3 - t^3 - 3st(s + t)

    6. express 2 sqrt 18 + sqrt 12 / sqrt 18 + 3 sqrt(12) 2 and 3 is multiplying

    7. if f:x ~~> 2/x, find the value of f(x + a) - f(x) / a

    i know this is a lot of questions, but my math teacher is extremely unhelpful and she tells us to do questions that she never taught before. i have no idea how to even start off for those problems so id prefer if someone can give the final answers also rather than hints etc cuz i don't think i can do them on my own. They look complicated and we've never done them in class and i don't know the rules regarding exponets etc.
    Last edited by mahi123; October 6th 2007 at 07:40 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mahi123 View Post
    1. x^1/3 = 32/ sqrt x 1/3 is the exponent

    2. if 9^x - 9^x-1 = 216, find 4^x x-1 is the exponent.
    Perhaps they are a review of stuff you learned in another class?

    Anyway:
    a^n \cdot a^m = a^{n + m}

    (a^n)^m = a^{nm}

    \frac{1}{a} = a^{-1}

    \sqrt[n]{a} = a^{1/n}
    (where a is a positive number, etc.)

    So problem 1:
    x^{1/3} = \frac{32}{\sqrt{x^{1/3}}}

    x^{1/3} = \frac{32}{(x^{1/3})^{1/2}}

    x^{1/3} = \frac{32}{x^{1/6}}

    x^{1/3} \cdot x^{1/6} = 32

    x^{1/3 + 1/6} = 32

    x^{2/3} = 32

    (x^{2/3})^{3/2} = 32^{3/2}

    Note that 32 = 2^5, so

    x = (2^5)^{3/2} = 2^{15/2}

    This can be simplified a little if you like. Note that 15/2 = 7 + 1/2. thus
    x = 2^{15/2} = 2^7 \cdot 2^{1/2} = 128 \sqrt{2}

    2)
    9^x - 9^{x-1} = 216

    Note that you can factor a 9^{x - 1} from the LHS:
    9^{x - 1}(9 - 1) = 216

    9^{x - 1} = \frac{216}{8} = 27

    Now, 27 is not a power of 9. But it is a power of 3:
    (3^2)^{x - 1} = 3^3

    3^{2x - 2} = 3^3

    The bases on each side of the equation are the same, so the exponents must be the same, too.
    2x - 2 = 3

    x = \frac{5}{2}

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Perhaps they are a review of stuff you learned in another class?
    thanks for the help, she says it's from gr 11 and i doubt every student remembers what they learnt in previous classes. I took it 2 years ago, but anyway, the thing is our math curriculum changed recently and before we had a course called functions in gr 11 and now they've made it advanced functions in gr 12, so whatever we did in gr 11 is supposed to be covered again in this course, it's part of the curriculum but she just takes unfair advantage of this and never goes over most of the stuff, we don't even have a textbook for this course( not made yet) so she just gives us one sheet and then we're on our own, and she gives no extra help either. this course is really important for me cuz it's gonna count towards my final ave for university , but i think she's gonna screw me up, i need atleast 85.
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    is up to his old tricks again! Jhevon's Avatar
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    Please use parentheses so we don't have to be guessing what you mean. if there are more than one terms in the denominator and/or numerator of a fraction, type it as (numerator)/(denominator) -- that is, with the parentheses
    Quote Originally Posted by mahi123 View Post

    3. find the inverse function of f(x) = 3sqrt (x + 1) ( on the left side of square root it has
    3 slightly on top so i guess it means whatever sqrt of x + 1 is it's to the power of 3.
    for the inverse, replace x with y and solve for y. that is, if y = \sqrt [3]{x + 1}, then, to find the inverse function, we set:

    x = \sqrt [3] {y + 1} and solve for y, and that would be the inverse function.



    4. rationalize the denominator

    5x/sqrt (x) - sqrt (2x-5)
    what is the denominator here? as you typed it, the only denominator is \sqrt {x}, the \sqrt {2x - 5} is a term unto itself, but i don't think that's what you meant. clarify

    5. prove that x-s-t is a factor of x^3 - s^3 - t^3 - 3st(s + t)
    note that x^3 - s^3 - t^3 - 3st(s + t) = x^3 - s^3 - t^3 - 3s^2t - 3st^2

    now, by the factor theorem (you must state this), if x - s - t is a factor, we must have that x = s + t is a root. thus if we plug in x = s + t, we should get zero, thus proving that x - s - t is a factor. try it and see what you get

    7. if f:x ~~> 2/x, find the value of f(x + a) - f(x) / a
    what you want is this:

    \frac {\frac 2{x + a} - \frac 2x}a

    now simplify


    Hint: begin by combining the fractions in the numerator
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    Quote Originally Posted by topsquark View Post

    2)
    9^x - 9^{x-1} = 216

    Note that you can factor a 9^{x - 1} from the LHS:
    9^{x - 1}(9 - 1) = 216

    9^{x - 1} = \frac{216}{8} = 27

    Now, 27 is not a power of 9. But it is a power of 3:
    (3^2)^{x - 1} = 3^3

    3^{2x - 2} = 3^3

    The bases on each side of the equation are the same, so the exponents must be the same, too.
    2x - 2 = 3

    x = \frac{5}{2}



    -Dan
    I understood everything except the part where you got 9^{x - 1} as a factor.

    Also in the first problem you wrote it wrong, i mean't 1/3 is the exponent on the left side, not that sqrt( x^1/3) , but i've got the answer for that so it's okay.
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  6. #6
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    Quote Originally Posted by Jhevon View Post



    what is the denominator here? as you typed it, the only denominator is \sqrt {x}, the \sqrt {2x - 5} is a term unto itself, but i don't think that's what you meant. clarify

    sqrt(2x-5) is also the denominator, sorry.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mahi123 View Post
    4. rationalize the denominator

    5x/sqrt (x) - sqrt (2x-5)
    Quote Originally Posted by mahi123 View Post
    sqrt(2x-5) is also the denominator, sorry.
    to rationalize the denominator, multiply by it's conjugate over itself and simplify:

    \frac {5x}{\sqrt{x} - \sqrt{2x - 5}} = \frac {5x}{\sqrt{x} - \sqrt{2x - 5}} \cdot \frac {\sqrt{x} + \sqrt{2x - 5}}{\sqrt{x} + \sqrt{2x - 5}}

    now simplify
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