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# If y is a function of x defined by ${a^{x + y}} = {a^x} + {a^y}$ where a is a real constant (a>1) then the domain of y(x) isA) $\left( {0, + \infty } \right)$B) $\left( { - \infty ,0} \right)$C) $\left( { - 1, + \infty } \right)$D) $\left( { - \infty ,1} \right)$ Verified
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Hint: To find the domain of ${a^{x + y}} = {a^x} + {a^y}$, first of all, simplify the function. Then we will take log on both sides and as we know that logarithm of a negative number is not possible, we will take the value of log greater than 0 and find the domain.

Complete step by step solution:
In this question, we are given that y is a function of x and is defined by ${a^{x + y}} = {a^x} + {a^y}$ and we need to find the domain of the function y(x).
Given: ${a^{x + y}} = {a^x} + {a^y}$
First of all, let us see the definition of range.
The domain of a function is the set of all possible input values that produce some output value range.
To find the domain of the given function, we need to simplify it further.
$\Rightarrow {a^{x + y}} = {a^x} + {a^y}$- - - - - -(1)
Now, we know the property that when two numbers having same base are multiplied, we add their powers. So we can write
${a^{x + y}} = {a^x} \cdot {a^y}$
Therefore, equation (1) becomes
$\Rightarrow {a^x} \cdot {a^y} = {a^x} + {a^y}$
Now, divide both LHS and RHS with ${a^x} \cdot {a^y}$, we get
$\Rightarrow \dfrac{{{a^x} \cdot {a^y}}}{{{a^x} \cdot {a^y}}} = \dfrac{{{a^x} + {a^y}}}{{{a^x} \cdot {a^y}}} \\ \Rightarrow 1 = \dfrac{1}{{{a^y}}} + \dfrac{1}{{{a^x}}} \\$
Now, we can write the inverse of any term as $\dfrac{1}{x} = {x^{ - 1}}$. Therefore,
$\Rightarrow 1 = {a^{ - x}} + {a^{ - y}} \\ \Rightarrow {a^{ - y}} = 1 - {a^{ - x}} \\$
Now, taking log with base a on both sides, we get
$\Rightarrow {\log _a}{a^{ - y}} = {\log _a}\left( {1 - {a^{ - x}}} \right) \\ \Rightarrow - y{\log _a}a = {\log _a}\left( {1 - {a^{ - x}}} \right) \\$
Now, we know that value of ${\log _a}a = 1$. Therefore, we get
$\Rightarrow - y = {\log _a}\left( {1 - {a^{ - x}}} \right) \\ \Rightarrow y = - {\log _a}\left( {1 - {a^{ - x}}} \right) \\$
Therefore,
$\Rightarrow 1 - {a^{ - x}} > 0 \\ \Rightarrow {a^{ - x}} < 1 \\ \Rightarrow {a^x} > 1 \\ \Rightarrow x > 0 \\$
Hence, the values can be anything greater than 0 to $\infty$.
Therefore, the domain of $y\left( x \right)$ is $\left( {0, + \infty } \right)$. So, our correct option is option (A).

Note:
Note that here we have taken $1 - {a^{ - x}} > 0$ because, logarithm of a negative number is not possible and hence the value of $1 - {a^{ - x}}$ must be greater than 0.
Other important point is that we have taken $x > 0$ for ${a^x} > 1$ as we know that ${a^x}$ will be equal to 1 only when $x = 0$, but here ${a^x} > 1$ and so the value of x cannot be 0 and hence, $x > 0$.