Calculate (4×7)^(-2): Negative Exponent Expression

Step 1: We begin by applying the power of a product rule to the expression $\left(4 \times 7\right)^{-2}$. According to this rule, $(ab)^n = a^n \times b^n$. Therefore, we have:

$\left(4 \times 7\right)^{-2} = 4^{-2} \times 7^{-2}$

Step 2: Next, we use the negative exponent rule, which states that $a^{-n} = \frac{1}{a^n}$. Applying this rule to both parts, we get:

$4^{-2} = \frac{1}{4^2}$ and $7^{-2} = \frac{1}{7^2}$

So, $4^{-2} \times 7^{-2} = \frac{1}{4^2} \times \frac{1}{7^2}$

By multiplying these fractions, we obtain:

$\frac{1}{4^2 \times 7^2}$

Therefore, the solution to the problem is $\frac{1}{4^2 \times 7^2}$.

Keep in mind - we could have used the rules in the other way around, first the negative exponent rule, and only then the product rule and the result would still be the same!