Simplify the Expression: 7² × (3⁵)⁻¹ × (1/4) × (1/3²)

First, let’s handle all the fractions in the problem by applying the law of negative exponents in reverse:

$\frac{1}{a^n} =a^{-n}$

Let's apply this law to the problem:

$7^2\cdot(3^5)^{-1}\cdot\frac{1}{4}\cdot\frac{1}{3^2}=7^2\cdot(3^5)^{-1}\cdot4^{-1}\cdot3^{-2}$

When we applied the above law of exponents to the third and fourth terms in the product—while leaving the first two terms unchanged—we did so in order to eliminate all fractions and make the expression easier to simplify.

Next, let’s recall the law of exponents for a power raised to a power:

$(a^m)^n=a^{m\cdot n}$

Let's apply this law to the second term in the multiplication we got in the last step:

$7^2\cdot(3^5)^{-1}\cdot4^{-1}\cdot3^{-2}=7^2\cdot3^{5\cdot(-1)}\cdot4^{-1}\cdot3^{-2}=7^2\cdot3^{-5}\cdot4^{-1}\cdot3^{-2}=7^2\cdot4^{-1}\cdot3^{-5}\cdot3^{-2}$

When in the first stage we applied the above law of exponents to the second term in the multiplication, then we simplified the expression and in the last stage we rearranged the expression using the commutative law of multiplication so that terms with identical bases are next to each other,

Let's continue and recall the law of exponents for multiplication of terms with identical bases:

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

Let's apply this law to the last term in the multiplication we got in the last step:

$7^2\cdot4^{-1}\cdot3^{-5}\cdot3^{-2}=7^2\cdot4^{-1}\cdot3^{-5+(-2)}=7^2\cdot4^{-1}\cdot3^{-5-2}=7^2\cdot4^{-1}\cdot3^{-7}$

We got the most simplified expression,

Let's summarize the solution steps so far, we got that:

$7^2\cdot(3^5)^{-1}\cdot\frac{1}{4}\cdot\frac{1}{3^2}=7^2\cdot3^{-5}\cdot4^{-1}\cdot3^{-2}=7^2\cdot4^{-1}\cdot3^{-5+(-2)}= 7^2\cdot4^{-1}\cdot3^{-7}$

Now let's note that there is no such answer among the given options, and another check of what we've done so far will reveal that there is no calculation error,

Therefore, we can conclude that additional mathematical manipulation is needed to determine which is the correct answer among the suggested answers,

Let's note that in answer A, the exponents of numbers 3 and 4 are identical to the exponents in the expression we got and these terms are indeed in the same place as the terms in the expression we got in the last step (meaning in the fraction's numerator (see important note at the end of the solution), both in answer A and in the expression we got), however, the difference between the expressions relates to the exponent of number 7 and its place in the fraction,

In our expression, the exponent is positive and it's in the numerator while in answer A its exponent is negative and it's in the denominator, which definitely reminds us of the negative exponent law:

$a^{-n}=\frac{1}{a^n}$

We'll continue in this direction and for simplicity, we'll deal separately with this term here and present it as a term with a negative exponent that's in the denominator:

$7^2=7^{-\underline {\bm{(-2)}}}=\frac{1}{7^{\underline {\bm{-2}}}}$

When in the first stage, in order to use the above law of exponents - we presented the term in question as having a negative exponent, while using the fact that:

$2=-(-2)$,

Then we applied the above law of exponents carefully, since the number that n represents in the above law of exponents in our use here is:

$-2$ (marked with an underline in the expression above)

Let's return to the expression we got and apply this understanding:

$7^2\cdot4^{-1}\cdot3^{-7}=7^{-(-2)}\cdot4^{-1}\cdot3^{-7}=\frac{1}{7^{-2}}\cdot4^{-1}\cdot3^{-7}=\frac{4^{-1}\cdot3^{-7}}{7^{-2}}$

When in the first through third stages we performed the mathematical manipulation we detailed earlier, and in the last stage we wrote the expression using one fraction line, while remembering that multiplication by a fraction means multiplication by the fraction's numerator,

Let's summarize the solution steps:

$7^2\cdot(3^5)^{-1}\cdot\frac{1}{4}\cdot\frac{1}{3^2}=7^2\cdot3^{-5}\cdot4^{-1}\cdot3^{-2}=7^2\cdot4^{-1}\cdot3^{-7}=\frac{4^{-1}\cdot3^{-7}}{7^{-2}}$

Therefore, the correct answer is indeed answer A.

Note:

When we say “the number in the numerator,” even if the expression is not written as a fraction, this is because any number can always be represented as a fraction with denominator 1. In other words, we can always write a number in the form:

$X=\frac{X}{1}$and therefore we can actually refer to $X$as a number that's in a fraction's numerator.