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Math Help - Need help solving few problems :(

  1. #1
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    Need help solving few problems :(

    Hello all, Im new to this forum and im not sure if this is the right place to post

    I need help in solving the following precalculus questions I do understand some of the steps but I always get lost and cannot solve them

    Can someone please help me solve them? I really need to do them for tomorrow I hope I'm not asking for much

    here's the link to the problems: http://i187.photobucket.com/albums/x...a/problems.jpg

    greets, amero
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  2. #2
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    1. I'll list the properties of logarithms and I'll do one as a model for the others.
    \ln{(A \cdot B)} = \ln{A} + \ln{B}

    \ln{\left(\frac{A}{B}\right)} = \ln{A} - \ln{B}

    \ln{(A)^n} = n\ln{A}

    \sqrt[n]{A} = A^{\frac{1}{n}}



    b) \ln{x} - 3\ln{(x+1)}

    \ln{x} - \ln{(x+1)^3}

    \ln{\frac{x}{(x+1)^3}}

    ===============

    2. 2x^2(x-1)^{\frac{1}{2}} - 5(x-1)^{-\frac{1}{2}}

    The question said that you need to take a common factor of the smaller exponent. Obviously, you can see that x-1 is common, so take the one with the smaller exponent.

    (x-1)^{-\frac{1}{2}} (2x^2(x-1) - 5)

    Now simplify.

    ===============

    3. First, you need to combine the fractions in the numerator.

    \frac{2}{x} - \frac{2}{x+1}

    Their Least Common Denominator is x(x+1). Here's how you combine them:

    \frac{2}{x} \cdot \frac{x+1}{x+1} - \frac{2}{x+1} \cdot \frac{x}{x}

    = \frac{2x + 2}{x^2 + x} - \frac{2x}{x^2 + x}

    = \frac{2x + 2 - 2x}{x^2 + x} = \frac{2}{x^2 + x}

    Can you continue?

    ===============

    4. Take a look at this post.
    http://www.mathhelpforum.com/math-he...980-post2.html

    ===============

    5 & 6. I'll list the properties of exponents just in case you don't remember them:

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

    \sqrt[n]{a} = a^{\frac{1}{n}}

    (a^{\frac{1}{n}})^n = a^{\frac{n}{n}} = a^1 = a

    a^0 = 1

    a^{-n} = \frac{1}{a^n}
    Also, remember the principal of powers:

    If a = b, then a^n = b^n.

    Alrighty now? Let's go back to the original question:

    4x^{\frac{2}{3}} = (30x + 4)^{\frac{1}{3}}

    Now, let's get rid of the fractional exponent by cubing both sides.

    ((4x)^{\frac{2}{3}})^3 = ((30x + 4)^{\frac{1}{3}})^3

    (4x)^2 = (30x + 4)^1

    16x^2 = 30x + 4

    So this becomes a quadratic form:

    16x^2 - 30x - 4 = 0

    Now, we have to solve for x. Use the quadratic formula:

    x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Now, for number 6: look again at the properties of exponents. What did I say a^0 equals to?
    Last edited by Chop Suey; August 14th 2008 at 06:36 AM.
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  3. #3
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    thx a lot man!

    now i just need 3 more that I'm also stuck on

    for the problem 1. a) i get the answer 12 / 5a .. is that correct?
    Can anyone please help

    http://i187.photobucket.com/albums/x...problems-1.jpg

    (I'm sorry that i give it in link but i have no idea how to write in LAteX )

    greets and again I'm sorry if I'm asking for much :P
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  4. #4
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    Hi Amero!
    You didn't get problem 1a) correct. Let's remember some properties of radicals quickly (it's actually properties of exponents, but for fractional exponents):

    \sqrt[n]{A \cdot B} = \sqrt[n]{A} \cdot \sqrt[n]{B}

    \sqrt[n]{\frac{A}{B}} = \frac{\sqrt[n]{A}}{\sqrt[n]{B}}

    Back to 1.a)

    \sqrt{\frac{48a}{20a^4}}

    = \frac{\sqrt{48a}}{\sqrt{20a^4}}

    = \frac{\sqrt{48}\sqrt{a}}{\sqrt{20}\sqrt{a^4}}

    = \frac{\sqrt{3 \cdot 16}\sqrt{a}}{\sqrt{4 \cdot 5}\sqrt{a^2 \cdot a^2}}

    = \frac{4\sqrt{3a}}{2a^2\sqrt{5}}

    Use this as a reference for question 1b).

    2. \frac{7}{2+\sqrt{7}}

    To simplify this by rationalizing the denominator, we multiply by the conjugate on both sides of the fraction. What this means is that we're going to use the conjugate of 2+\sqrt{7} to multiply by 1. The conjugate of 2 + \sqrt{7} is 2 - \sqrt{7}. Notice how I switched the signs.

    \frac{7}{2+\sqrt{7}} \cdot \frac{2-\sqrt{7}}{2-\sqrt{7}}


    The reason why people used to the do this before is because it was easier to find the value of the fraction then. See, when we multiply the denominator by the conjugate it becomes a difference between 2 squares.

    3. \frac{2x}{7} - \frac{1}{2} = \frac{3x+1}{2}

    Simply multiply both sides of the equation by ( 7\cdot 2):

    (7 \cdot 2) \left(\frac{2x}{7} - \frac{1}{2}\right) = (7\cdot 2) \left(\frac{3x+1}{2}\right)

    2(2x) - 7 = 7(3x+1)

    ...

    4. To write it in standard form, find the conjugate of the denominator and use it to multiply the given fraction by 1.

    P.S. If you want to learn LaTeX: http://www.mathhelpforum.com/math-he...-tutorial.html
    Last edited by Chop Suey; August 14th 2008 at 10:36 AM.
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by Chop Suey View Post
    Hi Amero!
    You didn't get problem 1a) correct. Let's remember some properties of radicals quickly (it's actually properties of exponents, but for fractional exponents):

    \sqrt[n]{A \cdot B} = \sqrt[n]{A} \cdot \sqrt[n]{B}

    \sqrt[n]{\frac{A}{B}} = \frac{\sqrt[n]{A}}{\sqrt[n]{B}}

    Back to 1.a)

    \sqrt{\frac{48a}{20a^4}}

    = \frac{\sqrt{48a}}{\sqrt{20a^4}}

    = \frac{\sqrt{48}\sqrt{a}}{\sqrt{20}\sqrt{a^4}}

    = \frac{\sqrt{3 \cdot 16}\sqrt{a}}{\sqrt{4 \cdot 5}\sqrt{a^2 \cdot a^2}}

    = \frac{4\sqrt{3a}}{2a{\color{red}^2}\sqrt{5}}
    Typo corrected in red above.
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  6. #6
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    Thank you Masters!

    Note to OP: if there is a radical in the denominator of a fraction, don't forget to rationalize it.
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