Math logs
posted by Mike .
Solve the system of equations:
y= (lnx)^2 + 2 (lnx^2)
y = 3ln(1/x^2)+24
Im not sure what to do..

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
(t6)(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