(e^x + e^-x)/2=1
To solve the equation (e^x + e^(-x))/2 = 1, we can follow these steps:
Step 1: Multiply both sides of the equation by 2 to get rid of the fraction:
2 * [(e^x + e^(-x))/2] = 2 * 1
e^x + e^(-x) = 2
Step 2: Multiply both sides of the equation by e^x to eliminate the negative exponent:
e^x * (e^x + e^(-x)) = e^x * 2
[e^(2x) + 1] = 2e^x
Step 3: Rearrange the equation:
e^(2x) - 2e^x + 1 = 0
Now we have a quadratic equation in terms of e^x. Let's make a substitution to simplify the equation:
Let y = e^x
The equation becomes:
y^2 - 2y + 1 = 0
This is a simple quadratic equation that can be factored as:
(y - 1)^2 = 0
Step 4: Solve for y:
(y - 1)^2 = 0
Taking the square root of both sides:
y - 1 = 0
y = 1
Step 5: Substitute back in for y:
e^x = 1
Step 6: Take the natural logarithm (ln) of both sides to solve for x:
ln(e^x) = ln(1)
x = ln(1)
The natural logarithm of 1 is 0, so the solution to the equation is x = 0. Therefore, when x = 0, the equation (e^x + e^(-x))/2 = 1 is satisfied.