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

Hint:
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.