A cure is represented parametrically by the equations x=f(t)=aIn⁡bt and y=g(t)=b−In⁡ata,b>0 and a≠1,b≠1 where t∈R

Solution:

Given the parametric equations \( x = f(t) = a \ln(bt) \) and \( y = g(t) = b - \ln(at) \), where \( a, b > 0 \) and \( a \neq 1, b \neq 1 \), \( t \in \mathbb{R} \):

1. Find the expression for \( t \) from \( x \):

\( x = a \ln(bt) \)

Solve for \( t \):

\( t = \frac{e^{x/a}}{b} \)

2. Substitute \( t \) in the expression for \( y \):

\( y = b - \ln(at) \)

Substitute \( t = \frac{e^{x/a}}{b} \):

\( y = b - \ln \left( a \cdot \frac{e^{x/a}}{b} \right) \)

\( y = b - \ln(a) - \ln \left( \frac{e^{x/a}}{b} \right) \)

\( y = b - \ln(a) - \left( \frac{x}{a} - \ln(b) \right) \)

\( y = b - \ln(a) - \frac{x}{a} + \ln(b) \)

\( y = b + \ln(b) - \ln(a) - \frac{x}{a} \)

Therefore, the curve can be expressed as:

\( y = b + \ln(b) - \ln(a) - \frac{x}{a} \)