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Math

posted by on .

I am having trouble figuring out how to solve these logarithms. Could someone please help!

log2(log4x)=1 and

solve for x and y:
(1/2)^x+y= 16 logx-y8=-3

  • Math - ,

    log2(log4x)=1
    Rewrite as
    (log((log(x))/(log(4))))/(log(2)) = 1
    Divide both sides by 1/(log(2)):
    log((log(x))/(log(4))) = log(2)
    Cancel logarithms by taking exp of both sides:
    (log(x))/(log(4)) = 2
    Divide both sides by 1/(log(4)):
    log(x) = 2 log(4)
    Cancel logarithms by taking exp of both sides:
    x = 16

    You need to rewrite the 2nd with parentheses.

    As written it is not clear.
    (1/2)^x+y= 16 logx-y8=-3

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