Logarithm -maths

here's a logarithm question which i am not able to solve. i need the help of you guys out there.

express the following in the form ln x = ax + b and find a and b

(xe^x)^2 = 30e^-x

answer's a = -1.5, b = 1.7

here's what did...

2x ln xe = -x ln 30e

ln x + ln e = -x (ln 30 - ln e) / 2x

ln x + 1 = -x (ln 30 - 1) / 2x

after that i got stuck. ><

P.S the above workings may not be correct for i'm unsure of whether i should bring down the -x on the right hand side.

You went wrong on your first part

"2x ln xe = -x ln 30e"

That's not how logarthims work.

Ans:

(xe^x)^2 = 30e^-x
2 ln(xe^x) = ln (30e^-x)
2 ln x +2 ln (e^x) = ln 30 + ln (e^-x)
2 ln x + 2x = ln 30 - x

so
2 ln x = -3x + ln 30
ln x = -1.5x + 0.5 ln 30

so a = -1.5, b = 0.5 ln30 = 1.7