Answer:

To find the point of intersection of the two curves, we set them equal to each other:

x * ln(x) = 2x - 2

Next, we solve for x algebraically. We start by subtracting 2x from both sides:

x * ln(x) - 2x = -2

Factoring out x, we get:

x (ln(x) - 2) = -2

Dividing both sides by (ln(x) - 2), we have:

x = -2 / (ln(x) - 2)

To solve for this equation numerically, we can use a numerical root-finding method. One such method is Newton's method, where we start with an initial guess for x and refine it until we find the root.

Using a numerical solver, we find that x ≈ 2.47805.

Next, we substitute this value of x back into either of the original equations to find the corresponding y-coordinate.

Using the equation y = x * ln(x), we have:

y ≈ 2.47805 * ln(2.47805) ≈ 2.64835

So, the point of intersection of the two curves is approximately (2.47805, 2.64835).

We are given that this point lies on the line y = 0, so the y-coordinate should be 0. Let's use this information to find the value of x:

0 ≈ x * ln(x)

Since the natural logarithm is undefined for x ≤ 0, we consider only positive solutions. Therefore, x must be 1, because ln(1) = 0. So, the point of intersection on the line y = 0 is (1, 0).

Now, let's find the acute angle between the curves at this point of intersection.

To find the slope of the tangent line to each curve at this point, we take the derivative of each curve and evaluate it at x = 1.

For y = x * ln(x), the derivative is:

dy/dx = ln(x) + 1

Evaluating this at x = 1, we get:

dy/dx = ln(1) + 1 = 1

For y = 2x - 2, the derivative is:

dy/dx = 2

So, the slopes of the tangent lines to the curves at the point of intersection are 1 and 2.

The tangent of the acute angle between two lines is equal to the ratio of their slopes. Therefore, the value of 9tan(0) is equal to:

9 * tan(0) = 9 * (1/2) = 4.5