Consider the following. x = e4t, y = t + 7 Eliminate the parameter to find a Cartesian equation of the curve.

  y = 7 +ln(x)/4

Step-by-step explanation:

Maybe your parameterized equation is ...

  (x, y) = (e^(4t), t +7)

Using t = y-7, we can substitute for t in the expression for x:

  x = e^(4(y -7))

  ln(x) = 4(y -7)

  ln(x) +28 = 4y

  y = 7 +ln(x)/4

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step-by-step explanation:

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