Post a New Question

Math logs

posted by .

Solve the system of equations:

y= (lnx)^2 + 2 (lnx^2)

y = 3ln(1/x^2)+24


Im not sure what to do..

  • Math logs -

    so we equate the two
    (lnx)^2 + 2(lnx^2) = 3ln(1/x^2)+24

    now some preliminary calculations:
    (lnx^2) = 2lnx
    so 2(lnx^2) = 4lnx
    and
    ln(1/x^2) = ln1 - lnx^2
    = 0 - 2lnx = - 2lnx

    so our equation becomes
    (lnx)^2 + 4lnx = 6lnx + 24
    let lnx = t
    so we have
    t^2 - 2t - 24 = 0
    (t-6)(t+4) = 0
    t = 6 or t = -4
    then lnx = 6 or lnx = -4
    but the second of those of course is undefined, so
    lnx = 6
    x = e^6

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question