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# f ${{a}^{m}}.{{a}^{n}}={{a}^{mn}}$ , then $m\left( n-2 \right)+n\left( m-2 \right)$ isA. 1B. -1C. 0D. $\dfrac{1}{2}$

Last updated date: 17th Sep 2024
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Hint: To solve the given equation we use the rule of exponentiation which is as follows ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$. Then, we simplify the equation and substitute the value. When we solve the equation further we get our answer.

We have given that ${{a}^{m}}.{{a}^{n}}={{a}^{mn}}$ .
We have to find the value of $m\left( n-2 \right)+n\left( m-2 \right)$ .
As given ${{a}^{m}}.{{a}^{n}}={{a}^{mn}}$
Now, we know that according to the rule of exponents if the bases (numbers) are same and the operation between the numbers is multiplication, then the powers of these numbers are just added.
${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$
So, the given equation becomes ${{a}^{m+n}}={{a}^{mn}}$
So, when we compare powers of both side of the equation we get
$m+n=mn.............(i)$
Now, we have to find the value of $m\left( n-2 \right)+n\left( m-2 \right)$ .
When we simplify the equation we get
\begin{align} & m\left( n-2 \right)+n\left( m-2 \right) \\ & =mn-2m+mn-2n \\ & =2mn-2m-2n \\ \end{align}
Now, substituting the value from equation (i), we get
$\Rightarrow 2\left( m+n \right)-2m-2n$
Now, simplifying further, we get
\begin{align} & \Rightarrow 2m+2n-2m-2n \\ & =0 \\ \end{align}
The value of $m\left( n-2 \right)+n\left( m-2 \right)$ is $0$ .
So, the correct answer is “Option C”.

Note: To solve this type of question one must have the idea about the properties of exponentiation. Exponents are nothing special but the powers of a number. Exponent or power tells us how many times the number is repeated. To simplify the expressions we need to follow the rules of exponents.