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(C) Given $a^x + a^y = a$. Since $f: A \rightarrow A$, the domain and range are both $A$. For $y$ to be defined, we must have $a^y > 0$, which implies

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$(-\infty, 1)$

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(C) Given $a^x + a^y = a$. Since $f: A \rightarrow A$, the domain and range are both $A$. For $y$ to be defined, we must have $a^y > 0$, which implies $a - a^x > 0$, so $a^x Since $a > 1$, taking $\log_a$ on both sides gives $x Thus, the domain is $x \in (-\infty, 1)$. Since the range must also be $A$, we have $y = \log_a(a - a^x)$. As $x \rightarrow -\infty$, $a^x \rightarrow 0$, so $y \rightarrow \log_a(a) = 1$. As $x \rightarrow 1^-$, $a^x \rightarrow a^-$, so $a - a^x \rightarrow 0^+$, which means $y \rightarrow -\infty$. Thus, the range is $(-\infty, 1)$. Therefore, $A = (-\infty, 1)$.

Let $a > 1$ be a constant. If $f: A \rightarrow A$ and $(x, y) \in f$ satisfy $a^x + a^y = a$, then $A =$

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