Answer

Verified

375k+ views

**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, $x > 0$.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE