Express x=e^(-2t), y=4e^(3t) in the form y = f(x) by eliminating the parameter.

Answer 1

x = e^(-2t)
ln x = -2t
t = (-1/2)lnx

y = 4e^3t
ln y = ln4 + 3t
t = (lny - ln4)/3

so (-1/2)lnx = (lny - ln4)/3
times 6
-3lnx = 2lny - 2ln4

3lnx + 2lny = 2ln4
ln x^3 + ln y^2 = ln16
ln( (x^3)(y^2) ) = ln 16
x^3 y^2 = 16
y^2 = 16/x^3
y = ± 4/√(x^3)

f(x) = ± 4/√(x^3)

Answer 2

or, note that

e^3t = (e^(-2t))^(-3/2) = x^(-3/2)
so, you have 4x^(-3/2)

I think the -4/√(x^3) is an extraneous solution, since y>0 for all t