Question 1196470
**1. Determine the Equation of the Line**

* **Given Points:** (1, 1) and (2, 5) 
    * Where (x², lg(2y+1)) are the coordinates

* **Calculate Slope (m):**
   * m = (y2 - y1) / (x2² - x1²) 
   * m = (5 - 1) / (2² - 1²) = 4 / 3

* **Calculate y-intercept (c):**
   * Using point (1, 1): 
      * 1 = (4/3) * 1² + c 
      * c = 1 - 4/3 = -1/3

* **Equation of the line:**
   * lg(2y+1) = (4/3)x² - 1/3

**2. Find y in terms of x (Part a)**

* lg(2y+1) = (4/3)x² - 1/3
* 2y+1 = 10^[(4/3)x² - 1/3] 
* 2y = 10^[(4/3)x² - 1/3] - 1
* y = [10^[(4/3)x² - 1/3] - 1] / 2

**3. Find the value of y when x = √3/2 (Part b)**

* y = [10^[(4/3)*(√3/2)² - 1/3] - 1] / 2
* y = [10^[(4/3)*(3/4) - 1/3] - 1] / 2
* y = [10^(1 - 1/3) - 1] / 2
* y = [10^(2/3) - 1] / 2
* y ≈ 1.8208

**4. Find the value of x when y = 2 (Part c)**

* lg(2*2 + 1) = (4/3)x² - 1/3
* lg(5) = (4/3)x² - 1/3
* (4/3)x² = lg(5) + 1/3
* x² = 3/4 * (lg(5) + 1/3)
* x = ± √[3/4 * (lg(5) + 1/3)]
* x ≈ ± 0.8799

**Therefore:**

* **a) y = [10^[(4/3)x² - 1/3] - 1] / 2**
* **b) y ≈ 1.8208 when x = √3/2**
* **c) x ≈ ± 0.8799 when y = 2**