# Math logs

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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

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