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Math Help - Complex number and Logarithm help me

  1. #1
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    Complex number and Logarithm help me

    Solve for x and y real:
    a) 7 - 2yi = 13x + 3i

    b.) sin x + 1/2i = 1 + (cosy)i

    solve for x:
    a.) (log_4_(x+1))^2 - 3log_4_(x+1) - 4 = 0

    _4_ is base
    thanks!!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Solve for x and y real:
    a) 7 - 2yi = 13x + 3i
    rearrange:

    13x+2y i = 7 - 3 i,

    Now as x and y are real equate the real and imaginary terms on each side
    or the equality to get:

    13x=7, and 2y=-3,

    or x=7/13, and y=-3/2.

    RonL

    Please double check the algebra.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    b.) sin x + 1/2i = 1 + (cosy)i
    Like the last one:

    sin(x) - i cos(y)=1 - (1/2) i,

    so sin(x)=1, and cos(y)=1/2, which you
    should be able to solve yourself

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    solve for x:
    a.) (log_4_(x+1))^2 - 3log_4_(x+1) - 4 = 0

    _4_ is base
    put y=log_4 (x+1), then the equation is:

    y^2-3y-4=0

    which has roots y=4, y=-1

    So in the first case we have log_4 (x+1)=4, or:

    x+1=4^4

    x=4^4-1=256-1=255.

    In the second case we have log_4 *x+1)=-1, or

    x+1=4^-1,

    x=3/4.

    RonL

    Again check the algebra yourself.
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