The area bounded by the curves
`xsqrt3+y=2log_(e)(x-ysqrt3)-2log_(e)2, y=sqrt3x,`
`y=-(1)/(sqrt3)x+2,` is

Options

A. `2log_(e)2` sq. units
B. `2log_(e)2+1` sq. units
C. `2log_(e)2-1` sq. units
D. `4log_(e)2-1` sq. units

Explanation

To find the area bounded by the curves given in the question, we will follow these steps:

### Step 1: Identify the curves
The curves given are:
1. \( y = \sqrt{3}x \)
2. \( y = -\frac{1}{\sqrt{3}}x + 2 \)
3. The curve defined by \( x\sqrt{3} + y = 2\log_e(x - y\sqrt{3}) - 2\log_e(2) \)

### Step 2: Find the intersection points of the curves
To find the area bounded by the curves, we first need to find the points where these curves intersect.

1. Set \( y = \sqrt{3}x \) equal to \( y = -\frac{1}{\sqrt{3}}x + 2 \):
\[ \sqrt{3}x = -\frac{1}{\sqrt{3}}x + 2 \]
Multiplying through by \( \sqrt{3} \) to eliminate the fraction:
\[ 3x = -x + 2\sqrt{3} \]
Rearranging gives:
\[ 4x = 2\sqrt{3} \quad \Rightarrow \quad x = \frac{\sqrt{3}}{2} \]
Substitute \( x \) back into \( y = \sqrt{3}x \):
\[ y = \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} \]
So one intersection point is \( \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \).

2. Next, we need to find the intersection of the first curve with the third curve. This may require substituting \( y = \sqrt{3}x \) into the third equation and solving for \( x \).

### Step 3: Set up the integral for the area
Once we have the intersection points, we can set up the integral to find the area between the curves. The area \( A \) can be expressed as:
\[ A = \int_{a}^{b} (f(x) - g(x)) \, dx \]
where \( f(x) \) is the upper curve and \( g(x) \) is the lower curve between the limits \( a \) and \( b \).

### Step 4: Calculate the area
Assuming we have determined the limits \( a \) and \( b \) and the functions \( f(x) \) and \( g(x) \), we can compute the integral. For example, if we found the area between \( x = 1 \) and \( x = 2 \):
\[ A = \int_{1}^{2} (f(x) - g(x)) \, dx \]

### Step 5: Evaluate the integral
Using integration techniques, we can evaluate the integral. For example:
\[ \int (x \log x - x) \, dx \]
This can be solved using integration by parts.

### Step 6: Final area calculation
After evaluating the integral, substitute the limits and simplify to find the area.