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Math Help - Simple exponential equations?

  1. #1
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    Simple exponential equations?

    Solve the following equations:

    a) 3√(2*4^x) = 8^(1-x)
    I worked it out to give 2^(2x+1) = 2^(9-9x) and thus x = 8/11.
    Is that correct?

    b) 16^x = 2 (2^(2x-1)) + 2


    Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?

    Step-by-step help would be very much appreciated =)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by erika View Post
    Solve the following equations:

    a) 3√(2*4^x) = 8^(1-x)
    I worked it out to give 2^(2x+1) = 2^(9-9x) and thus x = 8/11.
    Is that correct?
    [2 * 4^x]^{1/3} = 8^{1 - x} <-- Cube both sides

    2 * 4^x = [8^{1 - x}]^3

    2 * 4^x = 8^{3(1 - x)}

    2 * [2^2]^x = [2^3]^{3(1 - x)}

    2*2^{2x} = 2^{9(1 - x)}

    2^{2x + 1} = 2^{9(1 - x)}

    Equating exponents:
    2x + 1 = 9(1 - x)

    2x + 1 = 9 - 9x

    11x = 8

    x = 8/11

    So yes, you are correct!

    -Dan
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by erika View Post
    Solve the following equations:

    a) 3√(2*4^x) = 8^(1-x)
    I worked it out to give 2^(2x+1) = 2^(9-9x) and thus x = 8/11.
    Is that correct?
    Yes, you are correct



    b) 16^x = 2 (2^(2x-1)) + 2


    Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
    16^x = 2 (2^(2x-1)) + 2
    => 2^4x = 2^(2x) + 2
    => 2^(2x)2 = 2^(2x) + 2
    => 2^(2x)2 - 2^(2x) - 2 = 0 .......notice that this is a quadratic in 2^(2x)
    => (2^(2x))^2 - (2^(2x)) - 2 = 0 .....if you are the type that will get confused, you can replace 2^(2x) with some variable, say x, and it will look like a regular quadratic.

    => ((2^(2x)) - 2)((2^(2x)) + 1) = 0
    => ((2^(2x)) - 2) = 0 or ((2^(2x)) + 1) = 0
    => 2^(2x) = 2 or 2^(2x) = -1
    => 2^(2x) = 2 since 2^(2x) cannot be -1
    equating the powers we get
    2x = 1
    => x = 1/2
    Last edited by Jhevon; April 9th 2007 at 12:49 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by erika View Post
    b) 16^x = 2 (2^(2x-1)) + 2
    16^{x} = 2 * (2^{2x - 1}) + 2

    [2^4]^{x} = 2^{2x} + 2

    2^{4x} = 2^{2x} + 2

    2^{4x} - 2^{2x} - 2 = 0

    [2^{2x}]^2 - [2^{2x}] - 2 = 0

    Define y = 2^{2x}. Then
    y^2 - y - 2 = 0

    (y - 2)(y + 1) = 0

    Thus y = 2 or y = -1.

    So
    2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
    or
    2^{2x} = -1, which is impossible.

    Thus x = 1/2.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by erika View Post
    Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
    In general, you don't. However there are occasional special cases where you can. For example:
    Solve for x:
    (2^x)*(3^x) = 216

    (2*3)^x = 216

    6^x = 6^3

    Thus x = 3.

    So keep an eye out for ones you can do.

    -Dan
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