(e^x + e^-x)/2=1
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let e^x = y
y + 1/y = 2
y^2 - 2y + 1 = 0
(y-1)^2 = 0
y = 1
so e^x = 1 ----> x = 0
To solve the equation (e^x + e^-x)/2 = 1, we need to isolate the variable x. Let's go step by step.
Step 1: Multiply both sides of the equation by 2 to get rid of the fraction. This gives us:
e^x + e^-x = 2
Step 2: Multiply both sides of the equation by e^x to eliminate the exponent on the second term. This results in:
(e^x)^2 + 1 = 2e^x
Simplifying further, we get:
e^(2x) + 1 = 2e^x
Step 3: Subtract 2e^x from both sides of the equation to have all terms on one side:
e^(2x) - 2e^x + 1 = 0
Step 4: Rearrange the equation to form a quadratic equation by treating e^x as a variable:
(e^x)^2 - 2e^x + 1 = 0
Step 5: Recognize that the equation is in the form of a quadratic equation (ax^2 + bx + c = 0) with a = 1, b = -2, and c = 1.
Step 6: Solve the quadratic equation by factoring or using the quadratic formula:
Using factoring: The equation e^(2x) - 2e^x + 1 = 0 can be factored as:
(e^x - 1)(e^x - 1) = 0
Thus, e^x - 1 = 0, which implies e^x = 1. Taking the natural logarithm of both sides, we get:
x = ln(1) = 0
Using the quadratic formula: The quadratic formula states that for ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac))/(2a)
Applying this formula to our equation, we have:
x = (-(-2) ± √((-2)^2 - 4(1)(1)))/(2(1))
x = (2 ± √(4 - 4))/(2)
x = (2 ± √0)/2
Since the term inside the square root is 0, there is only one solution:
x = 2/2
x = 1
Therefore, the solution to the equation (e^x + e^-x)/2 = 1 is x = 0 and x = 1.