Let `f(x)=a^x-xlna`, a>1. Then the complete set of real values of x for which `f'(x)>0` is

Options

A. `(1,oo)`
B. `(-1,oo)`
C. `(0,oo)`
D. `(0,1)`

Answer

To find the complete set of real values of \( x \) for which \( f'(x) > 0 \) for the function \( f(x) = a^x - x \ln a \) where \( a > 1 \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(a^x) - \frac{d}{dx}(x \ln a) \]
Using the differentiation rules, we know:
- The derivative of \( a^x \) is \( a^x \ln a \).
- The derivative of \( x \ln a \) is \( \ln a \).

Thus, we have:
\[ f'(x) = a^x \ln a - \ln a \]

### Step 2: Factor out \( \ln a \)
We can factor out \( \ln a \) from the derivative:
\[ f'(x) = \ln a (a^x - 1) \]

### Step 3: Set the inequality \( f'(x) > 0 \)
For \( f'(x) \) to be greater than 0, we need:
\[ \ln a (a^x - 1) > 0 \]

### Step 4: Analyze the conditions
Since \( a > 1 \), we know that \( \ln a > 0 \). Therefore, the inequality simplifies to:
\[ a^x - 1 > 0 \]
This implies:
\[ a^x > 1 \]

### Step 5: Solve the inequality
Taking the logarithm of both sides:
\[ \ln(a^x) > \ln(1) \]
Since \( \ln(1) = 0 \), we have:
\[ x \ln a > 0 \]

### Step 6: Determine the sign of \( x \)
Since \( \ln a > 0 \) (because \( a > 1 \)), we can divide both sides by \( \ln a \) without changing the inequality:
\[ x > 0 \]

### Conclusion
The complete set of real values of \( x \) for which \( f'(x) > 0 \) is:
\[ x \in (0, \infty) \]