If \[f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)\]and \[f\left( 5 \right) = 32\;\;\] then \[f\left( {7} \right)\] is equal to
A) \[35\]
B) \[36\]
C) \[\dfrac{7}{5}\]
D) \[128\]

Hint: A function can be explained as a rule which takes each member x of a set and assigns, or maps it to an equivalent value of y known at its image.
x → Function → y
In this problem we will substitute the value of x and y till we get \[x + y = 5\]. Then try to find the value of \[f\left( {1} \right)\]. Then using this value we will find the value of \[f\left( {7} \right)\].

Complete step-by-step answer:
We have been given in the problem,
\[f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)\] and \[f\left( 5 \right) = 32\;\;\]
Taking \[x{\text{ }} = {\text{ }}y{\text{ }} = {\text{ }}1\] in above equation, we obtain
\[f\left( {1 + 1} \right){\text{ = }}f\left( 2 \right) = {\text{ }}f\left( 1 \right)f\left( 1 \right) = {\left[ {f(1)} \right]^2}\]
Similarly,
Taking \[x{\text{ }} = {\text{ 2}}\] and \[y{\text{ }} = {\text{ }}1\] in above equation, we obtain
\[f\left( {2 + 1} \right){\text{ = }}f\left( 3 \right) = {\text{ }}f\left( 2 \right)f\left( 1 \right) = {\left[ {f(1)} \right]^3}\]
Taking \[x{\text{ }} = {\text{ 2}}\] and \[y{\text{ }} = {\text{ 2}}\] in above equation, we obtain
\[f\left( {2 + 2} \right){\text{ = }}f\left( 4 \right) = {\text{ }}f\left( 2 \right)f\left( 2 \right) = {\left[ {f(1)} \right]^4}\]
Taking \[x{\text{ }} = {\text{ 2}}\] and \[y{\text{ }} = {\text{ 3}}\] in above equation, we obtain
\[f\left( {2 + 3} \right){\text{ = }}f\left( 5 \right) = {\text{ }}f\left( 2 \right)f\left( 3 \right) = {\left[ {f(1)} \right]^5}\]
Now replacing the value of $f(5)$ from\[f\left( 5 \right) = 32\;\;\] in the above equation, we get:
\[ \Rightarrow f\left( 5 \right) = 32\;\; = {\left[ {f(1)} \right]^5}\]
\[ \Rightarrow {\left[ {f(1)} \right]^5} = {2^5}\]
\[ \Rightarrow f(1) = 2\]
We have to find the value of \[f\left( {7} \right)\].
As we know 7= 2+5 and we already have a value of $f(5)$ so we will just calculate the value of $f(2)$
As we know \[{\text{ }}f\left( 2 \right) = {\left[ {f(1)} \right]^2}\]
$ \Rightarrow f(2) = {2^2} = 4$
Hence, for finding the value of \[f\left( {7} \right)\] using the equation given \[f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)\] we get
$ \Rightarrow f(7) = f(5 + 2) = f(5) \times f(2)$
Now substituting the value of $f(5)$ and $f(2)$we get:
$ \Rightarrow f(7) = 32 \times 4 = 128$
So, If \[f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)\] and \[f\left( 5 \right) = 32\;\;\]then \[f\left( {7} \right)\] is equal to \[128\].

Therefore, the option (D) is the correct answer.

Note: We can solve the question by second method in which we will keep on calculation the values of $f(2),f(3),f(4),f(5),f(6)$ and \[f\left( {7} \right)\] in terms of $f(1)$.
Taking \[x{\text{ }} = {\text{ 3}}\] and \[y{\text{ }} = {\text{ 3}}\] in above equation, we obtain
\[f\left( {3 + 3} \right){\text{ = }}f\left( 6 \right) = {\text{ }}f\left( 3 \right)f\left( 3 \right) = {\left[ {f(1)} \right]^6}\]
Taking \[x{\text{ }} = {\text{ 4}}\] and \[y{\text{ }} = {\text{ 3}}\] in above equation, we obtain
\[f\left( {4 + 3} \right){\text{ = }}f\left( 7 \right) = {\text{ }}f\left( 4 \right)f\left( 3 \right) = {\left[ {f(1)} \right]^7}\]
After this substituting the value of \[f(1) = 2\] in the above equation we get:
$ \Rightarrow f(7) = {2^7} = 128$