Find the exact value of the gradient of the curve y = e^(2- x)ln(3x- 2). at the point on the curve where x = 2.

To solve this problem, we must first differentiate:

Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2). 

Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2)

By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3e^(2- x)/(3x- 2) - e^(2- x)ln(3x- 2).

When we substitute x = 2 into this equation, we get that dy/dx = 3/4 - ln(4), which is our final answer.

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