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PreCalculus

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Hi i don't know how to do this problem

solve for n^3

given that log(40 SQRT(3))/log(4n) = log(45)/log(3n)

Thanks i'm copletely lost and don't know how to solve

  • PreCalculus -

    let a=log(40 sqrt(3)) and b=log45

    then

    a/log(4n)=b/log(3n)
    a(log3n)=b*log(4n)

    a(log3+logn)=b(log4+logn)

    logn(a-b)=blog4-alog3
    logn=( )/(a-b)

    logn^3=3logn= 3*above
    then take the antilog of both sides, it looks like fun.

  • RESPONSE -

    hmmm
    how did you go from this
    a(log3+logn)=b(log4+logn)
    to this?
    logn(a-b)=blog4-alog3
    logn=( )/(a-b)

  • PreCalculus -

    You are supposed to fill in the blanks.

    alog3+alogn=blon4=blogn
    logn(a-b)=blog4-alog3
    logn= (blog4-alog3)/(a-b)

  • RESPONSE -

    well if follow now but don't know how to take the anti log of that mumbo jumbo

    n = 10^(3((blog4-alog3)/(a-b) )

    im unsure how that simplifies

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