Logarithm question help needed

Express the following in the form ln x = ax + b and find the value of a and of b.
(a) xe^-x = 2.46
(b) (xe^x)^2 = 30e^-x

need the workings please

(a)
xe^-x = 2.46
ln (xe^-x) = ln (2.46), taking ln of both sides
ln x + ln(e^-x) = ln (2.46), using the law ln (rs) = ln r + ln s
ln x + (-x) = ln (2.46), using the fact that ln and exponential (base e) undo each other
ln x = x + ln (2.46), from adding x to both sides

So a = 1 and b = ln (2.46) in the ln x = ax + b form.

(b)
(xe^x)^2 = 30e^-x
ln [(xe^x)^2] = ln (30e^-x), taking ln of both sides
ln [(xe^x)^2] = ln 30 + ln (e^-x), using the law ln (rs) = ln r + ln s
2 ln (xe^x) = ln 30 + ln (e^-x), using the law ln (r^t) = t ln r
2 [ln x + ln(e^x)] = ln 30 + ln (e^-x), using the law ln (rs) = ln r + ln s
2 (ln x + x) = ln 30 + (-x), using the fact that ln and exponential (base e) undo each other
ln x + x = (1/2)(ln 30) - x/2, dividing both sides by 2
ln x = -3x/2 + (1/2)(ln 30) , subtracting x from both sides

So a = -3/2 and b = (1/2)(ln 30) in the ln x = ax + b form.

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