sum of all the values of x satisfying maximum `{e^x,e^(-x)}` = 4 is

Answer

To solve the problem of finding the sum of all values of \( x \) satisfying \( \max(e^x, e^{-x}) = 4 \), we will follow these steps:

### Step 1: Understand the function
The function we are dealing with is \( y = \max(e^x, e^{-x}) \). This function takes the maximum value between \( e^x \) and \( e^{-x} \).

### Step 2: Analyze the behavior of \( e^x \) and \( e^{-x} \)
- For \( x < 0 \): \( e^x < e^{-x} \) (since \( e^{-x} \) grows larger as \( x \) becomes more negative).
- For \( x = 0 \): \( e^x = e^{-x} = 1 \).
- For \( x > 0 \): \( e^x > e^{-x} \) (since \( e^x \) grows larger as \( x \) becomes positive).

### Step 3: Define the piecewise function
From the analysis:
\[ y = \begin{cases} e^{-x} & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ e^x & \text{if } x > 0 \end{cases} \]

### Step 4: Set up the equation
We need to solve the equation:
\[ \max(e^x, e^{-x}) = 4 \]

### Step 5: Solve for \( x < 0 \)
For \( x < 0 \):
\[ e^{-x} = 4 \]
Taking the natural logarithm of both sides:
\[ -x = \ln(4) \implies x = -\ln(4) \]

### Step 6: Solve for \( x = 0 \)
At \( x = 0 \):
\[ y = 1 \quad (\text{not equal to } 4) \]

### Step 7: Solve for \( x > 0 \)
For \( x > 0 \):
\[ e^x = 4 \]
Taking the natural logarithm of both sides:
\[ x = \ln(4) \]

### Step 8: Identify the solutions
The solutions we found are:
1. \( x = -\ln(4) \) (for \( x < 0 \))
2. \( x = \ln(4) \) (for \( x > 0 \))

### Step 9: Calculate the sum of all values of \( x \)
Now, we need to find the sum of all values of \( x \):
\[ \text{Sum} = -\ln(4) + \ln(4) = 0 \]

### Final Answer
The sum of all values of \( x \) satisfying \( \max(e^x, e^{-x}) = 4 \) is:
\[ \boxed{0} \]