Find the exact value of sinh(ln 2) and solve cosh2(x) = 1 for x.

You're on a roll — here are the solutions for the next problems: Problem 52: We need to find the exact value of sinh( 2). Step 1: Recall the definition of sinh(x). sinh(x) = e^x - e^-x2 Step 2: Substitute x = 2 into the definition of sinh(x). sinh( 2) = e^ 2 - e^- 22 Step 3: Simplify the exponential terms. Using the property e^ a = a: e^ 2 = 2 Using the property e^-a = (1)/(e^a): e^- 2 = e^(2^-1) = 2^-1 = (1)/(2) Step 4: Substitute these values back into the expression for sinh( 2). sinh( 2) = (2 - 1)/(2)2 Step 5: Calculate the value. First, simplify the numerator: 2 - (1)/(2) = (4)/(2) - (1)/(2) = (3)/(2). Then, divide by 2: sinh( 2) = (3)/(2)2 = (3)/(2) × (1)/(2) = (3)/(4) The final answer is (3)/(4). Problem 53: We need to find x given the equation ^2(x) = 1. Step 1: Solve for (x). Given ^2(x) = 1, taking the square root of both sides gives: (x) = ± 1 Step 2: Consider the range of (x). The range of the hyperbolic cosine function, (x), is [1, ). This means (x) can never be negative. Therefore, we must have: (x) = 1 Step 3: Recall the definition of (x). (x) = e^x + e^-x2 Step 4: Set the definition equal to 1 and solve for x. e^x + e^-x2 = 1 e^x + e^-x = 2 Step 5: Multiply the entire equation by e^x to eliminate the negative exponent. e^x(e^x) + e^x(e^-x) = 2e^x e^2x + e^0 = 2e^x e^2x + 1 = 2e^x Step 6: Rearrange the equation into a quadratic form. e^2x - 2e^x + 1 = 0 Step 7: Let y = e^x. Substitute y into the equation. y^2 - 2y + 1 = 0 Step 8: Solve the quadratic equation for y. This is a perfect square trinomial, which can be factored as (y-1)^2 = 0. (y-1)^2 = 0 y-1 = 0 y = 1 Step 9: Substitute back y = e^x and solve for x. e^x = 1 To solve for x, take the natural logarithm of both sides: (e^x) = (1) x = 0 The final answer is x=0. What's next?