# 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) is

A) $\left( {0, + \infty } \right)$

B) $\left( { - \infty ,0} \right)$

C) $\left( { - 1, + \infty } \right)$

D) $\left( { - \infty ,1} \right)$

Last updated date: 26th Mar 2023

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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, $x > 0$.

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