Negative Exponents: Using the laws of exponents

We use the formula:

$(\frac{a}{b})^{-n}=(\frac{b}{a})^n$

Therefore, we obtain:

$(\frac{3}{2})^4$

We use the formula:

$(\frac{b}{a})^n=\frac{b^n}{a^n}$

Therefore, we obtain:

$\frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}$

We begin by applying the power rule to the products within the parentheses:

$(z\cdot t)^n=z^n\cdot t^n$

That is, the power applied to a product within parentheses is applied to each of the terms when the parentheses are opened,

We apply the rule to the given problem:

$(8\cdot9\cdot5\cdot3)^{-2}=8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2}$

Therefore, the correct answer is option c.

Note:

Whilst it could be understood that the above power rule applies only to two terms of the product within parentheses, in reality, it is also valid for the power over a multiplication of multiple terms within parentheses, as was seen in the above problem.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms within parentheses (as formulated above), then it is also valid for a power over several terms of the product within parentheses (for example - three terms, etc.).

In order to solve this problem, we'll follow these steps:

• Step 1: Simplify each component using exponent rules

• Step 2: Apply multiplication and division of powers

• Step 3: Simplify the combined expression

Now, let's work through each step:

Step 1: Simplify $(\frac{1}{2})^8$. Using the power of a fraction rule, we have:

$\left(\frac{1}{2}\right)^8 = \frac{1^8}{2^8} = \frac{1}{2^8} = 2^{-8}$

Step 2: Substitute back into the original expression:

$\frac{2^{-4} \cdot 2^{-8} \cdot 2^{10}}{2^3}$

Combine the terms in the numerator using the product of powers rule:

$2^{-4} \cdot 2^{-8} \cdot 2^{10} = 2^{-4 + (-8) + 10} = 2^{-2}$

Now the expression becomes:

$\frac{2^{-2}}{2^3}$

Apply the division of powers rule:

$\frac{2^{-2}}{2^3} = 2^{-2 - 3} = 2^{-5}$

Thus, the solution to the problem is $2^{-5}$.

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.

Let's use the law of exponents for negative exponents:

$a^{-n} = \frac{1}{a^n}$and apply this law to the problem:

$10^8+10^{-4}+(\frac{1}{10})^{-16}=10^8+\frac{1}{10^4}+(10^{-1})^{-16}$$$when we apply the above law of exponents to the second term in the sum, and the same law but in the opposite direction - we'll apply it to the fraction inside the parentheses of the third term in the sum,

Now let's recall the law of exponents for exponent of an exponent:

$(a^m)^n=a^{m\cdot n}$we'll apply this law to the expression we got in the last step:

$10^8+\frac{1}{10^4}+(10^{-1})^{-16}=10^8+\frac{1}{10^4}+10^{(-1)\cdot(-16)}=10^8+\frac{1}{10^4}+10^{16}$when we apply this law to the third term from the left and then simplify the resulting expression,

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

$10^8+10^{-4}+(\frac{1}{10})^{-16}=10^8+\frac{1}{10^4}+(10^{-1})^{-16} =10^8+\frac{1}{10^4}+10^{16}$Therefore the correct answer is answer A.

First we will perform the multiplication of fractions using the rule for multiplying fractions:

$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

Let's apply this rule to the problem:

$3^x\cdot\frac{1}{3^{-x}}\cdot3^{2x}=\frac{3^x}{1}\cdot\frac{1}{3^{-x}}\cdot\frac{3^{2x}}{1}=\frac{3^x\cdot1\cdot3^{2x}}{1\cdot3^{-x}\cdot1}=\frac{3^x\cdot3^{2x}}{3^{-x}}$

In the first stage we performed the multiplication of fractions and then simplified the resulting expression,

Next let's recall the law of exponents for multiplication between terms with identical bases:

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

Let's apply this law to the numerator of the expression that we obtained in the last stage:

$\frac{3^x\cdot3^{2x}}{3^{-x}}=\frac{3^{x+2x}}{3^{-x}}=\frac{3^{3x}}{3^{-x}}$

Now let's recall the law of exponents for division between terms with identical bases:

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

Let's apply this law to the expression that we obtained in the last stage:

$\frac{3^{3x}}{3^{-x}}=3^{3x-(-x)}=3^{3x+x}=3^{4x}$

We applied the above law of exponents carefully, given that the term in the denominator has a negative exponent hence we used parentheses,

Let's summarize the solution so far:

$3^x\cdot\frac{1}{3^{-x}}\cdot3^{2x}=\frac{3^x\cdot3^{2x}}{3^{-x}} = \frac{3^{3x}}{3^{-x}}=3^{4x}$

Recall the law of exponents for power of a power but in the opposite direction:

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

Let's apply this law to the expression that we obtained in the last stage:

$3^{4x}=3^{4\cdot x}=\big(3^4\big)^x$

When we applied the above law of exponents instead of opening the parentheses and performing the multiplication between the exponents in the exponent (which is the direct way of the above law of exponents), we represented the expression in question as a term with an exponent in parentheses to which an exponent applies.

Therefore the correct answer is answer B.

We'll use the law of exponents for negative exponents:

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

Let's apply this law to the problem:

$5^4-(\frac{1}{5})^{-3}\cdot5^{-2}= 5^4-(5^{-1})^{-3}\cdot5^{-2}$

We apply the above law of exponents to the second term from the left.

Next, we'll recall the law of exponents for power of a power:

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

Let's apply this law to the expression that we obtained in the last step:

$5^4-(5^{-1})^{-3}\cdot5^{-2} = 5^4-5^{(-1)\cdot (-3)}\cdot5^{-2} = 5^4-5^{3}\cdot5^{-2}$

We apply the above law of exponents to the second term from the left and then simplify the resulting expression,

Let's continue and recall the law of exponents for multiplication of terms with the same base:

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

Let's apply this law to the expression that we obtained in the last step:

$5^4-5^{3}\cdot5^{-2} =5^4-5^{3+(-2)}=5^4-5^{3-2}=5^4-5^{1} =5^4-5$

We apply the above law of exponents to the second term from the left and then simplify the resulting expression,

From here, notice that we can factor the expression by taking out the common factor 5 from the parentheses:

$5^4-5 =5(5^3-1)$

Here we also used the law of exponents for multiplication of terms with the same base mentioned earlier, in the opposite direction:

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

Notice that:

$5^4=5\cdot 5^3$

Let's summarize the solution so far:

$5^4-(\frac{1}{5})^{-3}\cdot5^{-2}= 5^4-(5^{-1})^{-3}\cdot5^{-2} = 5^4-5^{3}\cdot5^{-2}=5(5^3-1)$

Therefore the correct answer is answer C.

To solve the problem, let's follow these steps:

• Step 1: Simplify the expression $45^{-80} \cdot \frac{1}{45^{-81}}$.
• Step 2: Simplify $49 \cdot 7^{-5}$.
• Step 3: Combine results to get the final expression.

Now, let's work through each step:
Step 1: Simplify $45^{-80} \cdot \frac{1}{45^{-81}}$. Using the exponent rule: $a^{-b} = \frac{1}{a^b}$ and $a^m \cdot a^n = a^{m+n}$, we have:

$45^{-80} \cdot \frac{1}{45^{-81}} = 45^{-80} \cdot 45^{81} = 45^{-80 + 81} = 45^1 = 45.$

Step 2: Simplify $49 \cdot 7^{-5}$. Note that $49 = 7^2$, so we can rewrite this as: $49 \cdot 7^{-5} = 7^2 \cdot 7^{-5} = 7^{2-5} = 7^{-3} = \frac{1}{7^3}.$

Step 3: Combine these results: $45 \cdot \frac{1}{7^3} = \frac{45}{7^3}.$

Therefore, the solution to the problem is $\frac{45}{7^3}$.

First let's note that the number 9 is a power of the number 3:

$9=3^2$

Therefore we can immediately move to a unified base in the problem, in addition we'll recall the law of powers for negative exponents but in the opposite direction:

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

Let's apply this to the problem:

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1}$

In the first term of the multiplication we replaced the number 9 with a power of 3, according to the relationship mentioned earlier, and simultaneously the third term in the multiplication we expressed as a term with a negative exponent according to the aforementioned law of exponents.

Now let's recall two additional laws of exponents:

a. The law of exponents for power of a power:

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

b. The law of exponents for multiplication between terms with equal bases:

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

Let's apply these two laws to the expression we got in the last stage:

$(3^2)^4\cdot3^{-8}\cdot3^{-1}=3^{2\cdot4}\cdot3^{-8}\cdot3^{-1}=3^8\cdot3^{-8}\cdot3^{-1}=3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

In the first stage we applied the law of exponents for power of a power mentioned in a', in the next stage we applied the law of exponents for multiplication of terms with identical bases mentioned in b', then we simplified the resulting expression.

Let's summarize the solution steps:

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1} =3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

Therefore the correct answer is answer b'.

Begin by writing the problem and converting the decimal fraction in the problem to a simple fraction:

$\frac{10^4\cdot0.1^{-3}\cdot10^{-8}}{1000}=\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000}=\text{?}$

Next

a. We'll use the law of exponents for negative exponents:

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

b. Note that the number 1000 is a power of the number 10:

$1000=10^3$

Apply the law of exponents from 'a' and the information from 'b' to the problem:

$\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000}=\frac{10^4\cdot(10^{-1})^{-3}\cdot10^{-8}}{10^3}$

We applied the law of exponents from 'a' to the term inside the parentheses of the middle term in the fraction's numerator. We applied the information from 'b' to the fraction's denominator,

Next, let's recall the law of exponents for power of a power:

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

And we'll apply this law to the same term we dealt with until now in the expression that we obtained in the last step:

$\frac{10^4\cdot(10^{-1})^{-3}\cdot10^{-8}}{10^3}=\frac{10^4\cdot10^{(-1)\cdot(-3)}\cdot10^{-8}}{10^3}=\frac{10^4\cdot10^3\cdot10^{-8}}{10^3}$

We applied the above law of exponents to the middle term in the numerator carefully, since the term in parentheses has a negative exponent. Hence we used parentheses and then proceeded to simplify the resulting expression,

Note that we can reduce the middle term in the fraction's numerator with the fraction's denominator. This is possible due to the fact that a multiplication operation exists between all terms in the fraction's numerator. Let's proceed to reduce:

$\frac{10^4\cdot10^3\cdot10^{-8}}{10^3}=10^4\cdot10^{-8}$

Let's summarize the various steps to our solution so far:

$\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000}=\frac{10^4\cdot(10^{-1})^{-3}\cdot10^{-8}}{10^3} = \frac{10^4\cdot10^3\cdot10^{-8}}{10^3} = 10^4\cdot10^{-8}$

Remember 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 expression that we obtained in the last step:

$10^4\cdot10^{-8}=10^{4+(-8)}10^{4-8}=10^{-4}$

Now let's once again apply the law of exponents for negative exponents mentioned in 'a' above:

$10^{-4}=\frac{1}{10^4}=\frac{1}{10000}=0.0001$

When in the third step we calculated the numerical result of raising 10 to the power of 4 in the fraction's denominator. In the next step we converted the simple fraction to a decimal fraction,

Let's summarize the various steps of our solution so far:

$\frac{10^4\cdot(\frac{1}{10})^{-3}\cdot10^{-8}}{1000} = \frac{10^4\cdot10^3\cdot10^{-8}}{10^3} = 10^{-4} =0.0001$

Therefore the correct answer is answer a.

First, we rewrite the given expression as a multiplication of fractions by applying the fraction multiplication rule in reverse:

$\frac{a\cdot c}{b\cdot d}= \frac{a}{b}\cdot\frac{c}{d}$

Let's apply this rule to the fraction in the problem:

$\frac{7^8}{7^{-4}\cdot4^2}\cdot32=\frac{7^8\cdot1}{7^{-4}\cdot4^2}\cdot32=\frac{7^8}{7^{-4}}\cdot\frac{1}{4^2}\cdot32$$$

In the first step, we recalled that any number can be written as itself times 1. In the next step, we rewrote the fraction as a product of separate fractions, ensuring that each fraction contained terms with identical bases.

Next, let's recall two laws of exponents:

a. The law of exponents for division between terms with identical bases:

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

b. The law of exponents for negative exponents but in reverse:

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

Let's apply these two laws of exponents to the expression we got in the last step:

$\frac{7^8}{7^{-4}}\cdot\frac{1}{4^2}\cdot32=7^{8-(-4)}\cdot4^{-2}\cdot32=7^{8+4}\cdot4^{-2}\cdot32=7^{12}\cdot4^{-2}\cdot32$

In the first step, we applied the law of exponents for division of terms with the same base (rule a) to the first factor, and the law of negative exponents (rule b) to the second factor. We then simplified the resulting expression.

Let's summarize the solution steps so far, we obtained the following:

$\frac{7^8}{7^{-4}\cdot4^2}\cdot32=\frac{7^8}{7^{-4}}\cdot\frac{1}{4^2}\cdot32=7^{8-(-4)}\cdot4^{-2}\cdot32=7^{12}\cdot4^{-2}\cdot32$

Now let's note an important fact:

The number 4 is a power of 2 and also the number 32 is a power of 2:

$4=2^2, \hspace{8pt}32=2^{5}$

Therefore, we can substitute these values into the expression from the previous step and rewrite it using base-2 terms:

$7^{12}\cdot4^{-2}\cdot32=7^{12}\cdot(2^2)^{-2}\cdot2^5$

From here we'll continue and recall two more laws of exponents:

c. The law of exponents for power of a power:

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

d. The law of exponents for multiplication between terms with identical bases:

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

Let's return to the problem and apply these two laws:

$7^{12}\cdot(2^2)^{-2}\cdot2^5=7^{12}\cdot2^{2\cdot(-2)}\cdot2^5=7^{12}\cdot2^{-4}\cdot2^5=7^{12}\cdot2^{-4+5}=7^{12}\cdot2^1=7^{12}\cdot2$

Where in the first step we applied the law of power of a power mentioned in c. to the second term in the multiplication, then we simplified the resulting expression and in the third step we applied the law of exponents for multiplication between terms with identical bases mentioned in d. and again simplified the resulting expression,

Let's summarize the solution steps so far, we obtained the following:

$\frac{7^8}{7^{-4}\cdot4^2}\cdot32=7^{12}\cdot4^{-2}\cdot32 =7^{12}\cdot(2^2)^{-2}\cdot2^5=7^{12}\cdot2^{2\cdot(-2)}\cdot2^5=7^{12}\cdot2$

Therefore using the multiplication substitution law we can observe that the correct answer is answer c.

First, let's note that in the problem there are terms with bases 2 and 3, and a term with base 81 which is inside the root,

Next, let's note that the number 81 is a power of the number 3:

$81=3^4$

Therefore we can replace it with this power of 3 in order to get a term with base 3, let's apply this in the problem:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{81}=\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}$

Next we want to get rid of the root in the problem so we'll recall the definition of the nth root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

Let's apply this law to the third term from the left in the product in the expression we got in the last step:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}=\frac{2^3}{3^2}\cdot3^{-2}\cdot \big(3^4\big)^{\frac{1}{4}}$

When we did this carefully since the term inside the root is a term with a power, therefore we used parentheses,

Next let's recall the power law for power of a power:

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

And let's apply this law in the expression we got in the last step on the same term in the product we dealt with until now:

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

When in the first stage we applied the above law on the third term in the product and then simplified the expression in the power exponent of that term while remembering that multiplication in a fraction means multiplication in the fraction's numerator,

Now let's recall the power law for negative power:

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

And we'll represent using it the last two terms in the product in the expression we got in the last step, as fractions, in order to later perform fraction multiplication:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot3=\frac{2^3}{3^2}\cdot\frac{1}{3^2}\cdot\frac{1}{3^{-1}}$

When for the second term from the left in the product we applied the above power law directly, and in the third term we applied it while understanding that we can represent the number 3 as a term with a negative power in the following way:

$$$3=3^1=3^{-(\underline{-1})}=\frac{1}{3^{\underline{-1}}}$

When we performed the use of the negative power law carefully, since the number that n represents in the above power law in our use here is:

$-1$

(Marked with underline in the expression above)

Let's summarize the solution steps until here, we got that:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}=\frac{2^3}{3^2}\cdot3^{-2}\cdot \big(3^4\big)^{\frac{1}{4}} =\frac{2^3}{3^2}\cdot3^{-2}\cdot3 = \frac{2^3}{3^2}\cdot\frac{1}{3^2}\cdot\frac{1}{3^{-1}}$

Next we'll perform the fraction multiplication while remembering the rule for multiplying fractions:

$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

Let's apply this rule in the last expression we got:

$\frac{2^3}{3^2}\cdot\frac{1}{3^2}\cdot\frac{1}{3^{-1}}=\frac{2^3\cdot1\cdot1}{3^2\cdot3^2\cdot3^{-1}}=\frac{2^3}{3^2\cdot3^2\cdot3^{-1}}$

Next let's recall the power law for multiplication between terms with equal bases:

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

And let's apply this law in the denominator of the expression we got in the last step:

$\frac{2^3}{3^2\cdot3^2\cdot3^{-1}}=\frac{2^3}{3^{2+2+(-1)}}=\frac{2^3}{3^{2+2-1}}=\frac{2^3}{3^3}$$$

When in the first stage we applied the above power law in the fraction's denominator and in the following stages we simplified the resulting expression,

Finally let's recall another important power law - it's the power law for power applied to parentheses:

$$$\big(\frac{a}{c}\big)^n=\frac{a^n}{c^n}$

And let's apply it in the expression we got in the last step:

$\frac{2^3}{3^3}=\big(\frac{2}{3}\big)^3$

When we applied the above power law in the opposite direction, meaning - instead of opening the parentheses and applying the power to the fraction's numerator and denominator, we used the fact that both the numerator and denominator are raised to the same power and therefore we can represent the expression as a fraction raised to a power, which can be done only because both the numerator and denominator are raised to the same power.

Let's summarize the solution steps until here, we got that:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}=\frac{2^3}{3^2}\cdot3^{-2}\cdot3 = \frac{2^3}{3^2\cdot3^2\cdot3^{-1}} =\frac{2^3}{3^3} = \big(\frac{2}{3}\big)^3$

Therefore the correct answer is answer A.