Simplify the Expression: a²×2a³×3a⁻¹ Using Exponent Rules

Note that there is multiplication between all terms in the expression, hence we'll apply the distributive property of multiplication to understand that we can handle the coefficients of terms raised to powers as well as the terms themselves separately. For clarity, let's handle it in steps:

$a^2\cdot2a^3\cdot3a^{-1}=2\cdot3\cdot a^2\cdot a^3\cdot a^{-1}=6\cdot a^2\cdot a^3\cdot a^{-1}$

Given that there is multiplication between all terms, we could do this. It should be noted that we can (and it's preferable) to skip the middle step, meaning:

Write directly:$a^2\cdot2a^3\cdot3a^{-1}=6\cdot a^2\cdot a^3\cdot a^{-1}$

From here on we will no longer write the multiplication sign and remember that it is conventional to simply place the terms next to each other\ place the term next to its coefficient to indicate multiplication between them,

Next apply the law of exponents for multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Note that this law applies to any number of terms being multiplied and not just two, for example for three terms with identical bases we obtain the following:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$

When we used the above law of exponents twice, we can also perform the same calculation for four terms in multiplication five and so on..,

Let's return to the problem and apply the above law of exponents:

$6a^2a^3a^{-1}=6a^{2+3-1}=6a^4$

Therefore the correct answer is C.

Important note:

Here we need to emphasize that we should always ask the question - what does the exponent apply to?

For example, in this problem the exponent applies only to the base of $a$ and not to the numbers, more clearly, in the following expression: $5b^7$The exponent applies only to $b$ and not to the number 5,

whereas when we write: $(5b)^7$The exponent applies to each of the multiplication terms inside the parentheses,

meaning: $(5b)^7=5^7b^7$

This is actually the application of the law of exponents:

$(x\cdot y)^n=x^n\cdot y^n$

Which follows both from the meaning of parentheses and from the definition of exponents.