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Math Help - Logarithms and surds

  1. #1
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    Logarithms and surds

    To be exact, I have a few questions on logarithms.. The questions are listed as follow. It would be great if you could show necessary workings and solution.

    Question 1
    Prove that the equation has a real root from the root x=2. Find this root.

    2^(2x) + 64[2^(-2)] =32

    Question 2
    Given that 2y = a^x + a^(-x) where a>1 and x>0
    Prove that a^x = y + (y^2 -1)^1/2

    Similarly if 2z = a^(3x) = a^(-3x) prove that z=4y^3 -3y

    Question 3
    Solve the equation
    5/(x+1)^1/2 -3 (x+1)^1/2 + 2 = 0
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  2. #2
    MHF Contributor Unknown008's Avatar
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    2.

    2y = a^x + a^(-x)

    2y = a^x + \frac{1}{a^x}

    Multiply by a^x throughout:

    2y(a^x) = a^x(a^x) + 1

    2y(a^x) = a^{2x} + 1

    0 = a^{2x} - 2y(a^x) + 1

    Therefore, completing the square;

    0 = (a^x - y)^2 - y^2 + 1

    Rearrange for a^x;

    y^2 - 1 = (a^x - y)^2

     (a^x - y) = (y^2 - 1)^{\frac12}

     a^x  = y + (y^2 - 1)^{\frac12}

    Ok, that's as far as I'm sure for this part. For the other one, I'm not quite getting something.

    If z = a^3x = a^-3x;

    a^x = z^1/3 = z^-1/3.

    Taking a^x = z^1/3;

    z^{1/3} = y + (y^2 - 1)^{\frac12}

    z = (y + (y^2 - 1)^{\frac12})^3

    Expanding, I get:

    z = 4y^3 -3y + (y^2-1)^{\frac12}(2y^2-1)

    Maybe it's because I didn't take a^x = z^-1/3, but if I do so, I'll get infinite terms, and I don't know how to deal with that.

    3. \frac{5}{(x+1)^{\frac12}} - 3(x+1)^{\frac12} + 2 = 0

    Is that it? If so, multiply throughout by (x+1)^1/2;

    5 - 3(x+1) + 2(x+1)^{\frac12} = 0

    You can make use of a substitution to help you solve here. Let (x+1)^1/2 = y;

    5 - 3y^2 + 2y = 0

    3y^2 - 2y - 5 =0

    (3y - 5)(y + 1)=0

    So, 3y -5 = 0,

    y = 5/3

    (x+1)^{\frac12} = \frac53

    and y +1 = 0

    y = -1

    (x+1)^{\frac12} = -1

    You can solve those two now.

    For the first part, I prefer not answering because I don't quite understand what the question is asking for. I can solve for x, and get x = 2, but not show that it's a real root, sorry
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  3. #3
    MHF Contributor
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    Hello everyone

    I'm sure the second part of number 2 should read:
    2z=a^{3x}+a^{-3x}
    So we can say:
    2z = (a^x)^3 + (a^{-x})^3
    = (a^x+a^{-x})^3 - 3(a^x)^2(a^{-x})-3(a^x)(a^{-x})^2, using (p+q)^3 = p^3+3p^2q+3pq^2+q^3

    = (a^x+a^{-x})^3 - 3\Big((a^x)+(a^{-x})\Big)


    =(2y)^3-3\cdot2y

    \Rightarrow z = 4y^3-3y
    Grandad
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