Find the least value of `k` such that the equation `(ln x) + k = e^(x-k)` has a solution.

Options

A. `e`
B. `1//e`
C. `1`
D. None of these

Explanation

To find the least value of \( k \) such that the equation

\[ \ln x + k = e^{(x - k)} \]

has a solution, we can follow these steps:

### Step 1: Rearranging the equation
We start by rearranging the given equation:

\[ \ln x + k - e^{(x - k)} = 0 \]

Let \( f(x) = \ln x + k - e^{(x - k)} \). We want to find \( k \) such that \( f(x) = 0 \) has at least one solution.

### Step 2: Analyzing the functions
We know that \( \ln x \) is defined for \( x > 0 \) and approaches \(-\infty\) as \( x \) approaches \( 0 \). The function \( e^{(x - k)} \) is an exponential function that is always positive and increases rapidly as \( x \) increases.

### Step 3: Finding the derivative
To find the conditions under which \( f(x) = 0 \) has a solution, we differentiate \( f(x) \):

\[ f'(x) = \frac{1}{x} - e^{(x - k)} \]

### Step 4: Setting the derivative to zero
Setting \( f'(x) = 0 \) gives us:

\[ \frac{1}{x} = e^{(x - k)} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{1}{x}\right) = x - k \]

This simplifies to:

\[ -k = x - \ln x \]

Thus, we have:

\[ k = x + \ln x \]

### Step 5: Finding the minimum value of \( k \)
To find the minimum value of \( k \), we need to analyze the function \( g(x) = x + \ln x \). We differentiate \( g(x) \):

\[ g'(x) = 1 + \frac{1}{x} \]

Since \( g'(x) > 0 \) for \( x > 0 \), \( g(x) \) is an increasing function. Therefore, the minimum value occurs as \( x \) approaches \( 0 \).

### Step 6: Evaluating the limit
As \( x \to 0^+ \):

\[ g(x) = x + \ln x \to 0 - \infty = -\infty \]

However, we need to find the least value of \( k \) such that \( f(x) = 0 \) has a solution.

### Step 7: Finding the intersection point
To find the least value of \( k \) such that the curves intersect, we can find the point where \( \ln x \) and \( e^{(x - k)} \) are tangent. This occurs when:

\[ \ln x = e^{(x - k)} \]

At the point of tangency, we can set \( k = 1 \) because:

- At \( x = 1 \), \( \ln(1) = 0 \) and \( e^{(1 - k)} = e^{(1 - 1)} = 1 \).
- Thus, \( k = 1 \) is the least value that allows for a solution.

### Conclusion
The least value of \( k \) such that the equation has a solution is:

\[ \boxed{1} \]