Final answer:
To find the equation of the tangent line to the curve ln(xy) = 2x at the point (1, e^2), differentiate the equation, substitute the given point, and use the point-slope form of a line.
Step-by-step explanation:
To find the equation of the tangent line to the curve ln(xy) = 2x at the point (1, e^2), we need to find the slope of the tangent line and the y-intercept.
First, we differentiate the equation ln(xy) = 2x with respect to x using the product rule. The derivative is (y + xy')/(xy) = 2.
Next, we substitute x = 1 and y = e^2 into the derivative to find the slope. The slope is equal to (e^2 + e^2y')/(e^2) = 2.
Finally, we use the point-slope form of a line with the slope of 2 and the point (1, e^2) to find the equation of the tangent line: y - e^2 = 2(x - 1).