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    i) show that log2w = 2log4w

    The above I know how to do, but:

    ii) Hence solve the simultaneous equations:

    log4t² - log2U² = (log0.59) - 1

    4^t = 2^(U-2)

    where U>0 and t>0

    Please help, thanks.
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    First we convert log[0.5](9) to base-2. (You can use a standard formula here if you prefer.) Let x = log[0.5](9). Then 0.5^x = 9, so 1 = 9*2^x, so 0 = log[2](9) + x. So log[0.5](9) = -log[2](9)

    Now we simplify the more complicated equation:

    log[4](t^2) - log[2](U^2) = log[0.5](9) - 1
    (1/2)log[2](t^2) - log[2](U^2) = -log[2](9) - 1 . . . . . converting everything to base-2
    log[2](t) - log[2](U^2) = -log[2](9) - 1
    log[2](U^2) - log[2](t) - log[2](9) = 1
    log[2](U^2/(9t)) = 1
    U^2/(9t) = 2
    U^2 = 18t

    Since also 4^t = 2^(U - 2),

    2^(2t) = 2^(U - 2)
    2t = U - 2
    (1/9)U^2 = U - 2
    U^2 - 9U + 18 = 0
    (U - 6)(U - 3) = 0
    U = 3 or 6

    So the solutions to the given equations are:

    (A) U = 3, t = 1/2;
    (B) U = 6, t = 2.

    That was quite hard.
 
 
 
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