Question

When multiplying exponents, such as \[x\] times \[x\] , would the answer be \[{x^2}\] or \[2x\] ?

Answer 1

Hint: We use the concept of exponents and basic variable algebra to solve this problem. We use addition, multiplication and some other operations of variables in this problem. A variable is a term which has no constant value and changes by time to time with conditions and situations. We also know that multiplication is repeated addition.

Complete step-by-step solution:
A variable can be represented by any letter like \[x,y,z,a,b,c,...\]
Terms containing these variables are also called variable terms. For example, terms like \[2x,5y,10a,....\] are also variable terms.
Terms with the same variable are called like terms and terms with different variables are called unlike terms.
For example, \[2x{\text{ and }}6x\] are like terms.
\[5ax{\text{ and }}\dfrac{{ax}}{8}\] are also like terms.
\[{\text{67y and }}\sqrt x \] are unlike terms.
According to variable algebra, like terms can be added or subtracted and unlike terms can’t be.
We know that multiplication is a repeating addition process.
That means, \[3 \times 4\] means that 3 is added to itself four times.
So, \[3 \times 4 = 3 + 3 + 3 + 3\]
And, if a number is multiplied to it for some number of times, then it forms an exponent.
Consider the example, \[5 \times 5 \times 5 \times 5\]
Here, 5 is multiplied to itself for four times, so it can be written as \[5 \times 5 \times 5 \times 5 = {5^4}\]
So, in the question, it is given as \[x\] times \[x\] . So, \[x\] is multiplied to itself for two times, so it can be written as \[x{\text{ times }}x = x \times x = {x^2}\]
And, here the variable is getting multiplied to itself but not getting added to itself.
So, \[x{\text{ times }}x\] should be written as \[{x^2}\] but not as \[2x\].

Note: We get an exponent form only when like terms are multiplied to themselves.
That means, \[a \times a \times a = {a^3}\] and \[p \times {q^2} = p{q^2}\]
We can also use identities like \[{a^m} \times {a^n} = {a^{(m + n)}}\]
So, we get \[x \times x = {x^1} \times {x^1} = {x^{(1 + 1)}} = {x^2}\]
We can also find the answer in this way.