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    log (base 2) x - log (base 4) y = 4
    log (base 2) (x - 2y) = 5.
    Solve the simultaneous equations? Good Luck
    can u write the steps cos i dont get the right answer!!!
    answer: x = 64, y = 16
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    Arrrrggggghhhhhh!!!!!!!!!!!!!!

    I think there must be an error in the question, or what you typed...I CAN'T get the answers you give...

    This is the rearrangement I do:

    log (base 2) (x - 2y) = 5
    x - 2y = 2^5
    x = 2^5 +2y OR y = 1/2x - 16

    I then do this with the other:

    log (base 2) x - log (base 4) y = 4
    2log (base 4) x - log (base 4) y = 4
    log (base 4) x^2 - log (base 4) y = 4
    x^2 - y = 4^4

    Substitute:

    x^2 - 1/2x + 16 = 256
    x^2 - 1/2x - 240 = 0

    However I can't get a nice round answer from any substitution of that nature... :confused:

    btw. In case you are confused about my rearrangement of the equation with the two bases; the trick is to use the fact that any power of 4 that gives "x" is half the power of 2 that gives that same "x", because 2^2 = 4.
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    here is your error

    >log (base 4) x^2 - log (base 4) y = 4
    >x^2 - y = 4^4

    it should be:
    log (base 4) x^2 - log (base 4) y = 4

    (x^2)/y = 4^4

    etc.......

    Dont you dare make that mistake tommorow....

    Mrm.
 
 
 
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