Question

The domain of the definition of the function $y\left( x \right)$ given by the equation ${a^x} + {a^y} = a$, where $a > 1$, is
(a) $0 < x \le 1$
(b) $0 \le x < 1$
(c) $- \infty < x < 1$
(d) $- \infty < x \le 0$

Answer 1

Here, we need to find the domain of the given function. We will rewrite the given equation using logarithms. Then, we will use the domain of a logarithmic function to find the interval where $x$ lies, and hence, find the correct option.

Complete step by step solution:
First, we will simplify the given equation.
Subtracting ${a^x}$ from both sides of the equation, we get
$\Rightarrow {a^x} + {a^y} - {a^x} = a - {a^x}$
Therefore, we get
$\Rightarrow {a^y} = a - {a^x}$
Rewriting the equation ${a^y} = a - {a^x}$ using logarithms, we get
${\log _a}\left( {a - {a^x}} \right) = y$
Now, we know that if ${\log _b}x = y$, then $x > 0$.
Therefore, since ${\log _a}\left( {a - {a^x}} \right) = y$, we get
$a - {a^x} > 0$
Adding ${a^x}$ on both sides of the inequation, we get
$\begin{array}{l} \Rightarrow a - {a^x} + {a^x} > 0 + {a^x}\\\ \Rightarrow a > {a^x}\end{array}$
Any number raised to the power 1 is equal to itself, that is ${x^1} = x$.
Substituting ${a^1}$ for $a$ in the equation $a > {a^x}$, we get
$\Rightarrow {a^1} > {a^x}$
The bases of the expressions on both sides are the same.
Therefore, from ${a^1} > {a^x}$, we get
$\begin{array}{l} \Rightarrow 1 > x\\\ \Rightarrow x < 1\end{array}$
$\therefore$ The value of $x$ is all real numbers less than 1.
Thus, the domain of the given function is
$\begin{array}{l} \Rightarrow x \in \left( { - \infty ,1} \right)\\\ \Rightarrow - \infty < x < 1\end{array}$
Therefore, the domain of the definition of the function $y\left( x \right)$ given by the equation ${a^x} + {a^y} = a$, where $a > 1$, is $- \infty < x < 1$.

The correct option is option (c).

Note:
We rewrote the equation ${a^y} = a - {a^x}$ as ${\log _a}\left( {a - {a^x}} \right) = y$ using logarithms. If an equation is of the form ${b^y} = x$, it can be written using logarithms as ${\log _b}x = y$, where $x > 0$, $b > 0$ and $b$ is not equal to 1.
We used the condition $x > 0$ to find the domain of the function.
We will check whether our answer meets the other conditions $b > 0$ and $b$ is not equal to 1.
Comparing ${\log _a}\left( {a - {a^x}} \right) = y$ and ${\log _b}x = y$, we can observe that
$b = a$
It is given that $a > 1$. Therefore, $a > 0$ and $a$ is not equal to 1.
Thus, we have verified the other conditions.