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December 21, 2014

December 21, 2014

Posted by **Mike** on Wednesday, October 14, 2009 at 12:53am.

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

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

Im not sure what to do..

- Math logs -
**Reiny**, Wednesday, October 14, 2009 at 8:28amso 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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