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    (a) Solve 2e^x + 2e^-x = 5

    (b) If xy = 64 and logx y (where x is base) + logy x (where y is base) = 5/2 find x and y.

    Also are we allowed the use of a mirror in the P2 exam??
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    (Original post by !Laxy!)
    (a) Solve 2e^x + 2e^-x = 5
    Multiplying by e^x gives:
    2e^2x-5e^x+2=0
    2(e^x)^2-5(e^x)+2=0
    You have a quadratic in e^x. Solve for e^x and hence find x.

    (b) If xy = 64 and logx y (where x is base) + logy x (where y is base) = 5/2 find x and y.
    y=64/x
    lny/lnx+lnx/lny=5/2
    (lny)^2+(lnx)^2=(5/2)(lny)(lnx)
    [ln(64/x)]^2+(lnx)^2=(5/2)(lnx)[ln(64/x)]
    [ln64-lnx]^2+[lnx]^2=(5/2)(lnx)(ln64-lnx)]
    If you multiply everything out you'll end up with a quadratic in (lnx) so you can solve to find lnx, x and hence the cooresponding value of y.
 
 
 
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