The range of the function y = f(x) if `(34)^(x) + (34)^(y) = 34` equals

Options

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

Answer

To find the range of the function given by the equation \(34^x + 34^y = 34\), we can follow these steps:

### Step 1: Rearranging the Equation
Start with the equation:
\[ 34^x + 34^y = 34 \]
We can rearrange this to isolate \(34^y\):
\[ 34^y = 34 - 34^x \]

### Step 2: Analyzing the Right Side
For the right side \(34 - 34^x\) to be positive (since \(34^y\) must be positive), we need:
\[ 34 - 34^x > 0 \]
This implies:
\[ 34 > 34^x \]
Dividing both sides by \(34\) gives:
\[ 1 > 34^{x-1} \]

### Step 3: Finding the Condition for \(x\)
The inequality \(1 > 34^{x-1}\) holds true when:
\[ x - 1 < 0 \quad \text{or} \quad x < 1 \]

### Step 4: Finding the Expression for \(y\)
Now, substituting back into the equation for \(y\):
\[ 34^y = 34 - 34^x \]
Taking logarithm base \(34\) on both sides:
\[ y = \log_{34}(34 - 34^x) \]

### Step 5: Finding the Range of \(y\)
Since \(x < 1\), we can analyze the behavior of \(34 - 34^x\):
- As \(x\) approaches \(1\), \(34^x\) approaches \(34\), making \(34 - 34^x\) approach \(0\).
- As \(x\) approaches \(-\infty\), \(34^x\) approaches \(0\), making \(34 - 34^x\) approach \(34\).

Thus, \(34 - 34^x\) varies from \(34\) (when \(x \to -\infty\)) to \(0\) (when \(x \to 1\)). Therefore, \(y\) will vary as follows:
- When \(34 - 34^x = 34\), \(y = \log_{34}(34) = 1\).
- When \(34 - 34^x \to 0\), \(y \to -\infty\).

### Conclusion
Thus, the range of \(y\) is:
\[ (-\infty, 1) \]

### Final Answer
The range of the function \(y = f(x)\) is \((- \infty, 1)\).

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