Power of a Quotient Rule for Exponents: Using multiple rules

We need to calculate division (fraction=division operation between numerator and denominator) between terms with identical bases, therefore we will use the law of exponents for division between terms with identical base:

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

Note that this law can only be used to calculate division between terms with identical bases.

In this problem, there is also a term with a negative exponent, but this does not pose an issue regarding the use of the aforementioned law of exponents. In fact, this law of exponents is valid in all cases for numerical terms with different exponents, including negative exponents, rational number exponents, and even irrational number exponents, etc.

Let's return to the problem and apply the aforementioned law of exponents for each fraction separately:

$\frac{y^3}{y^6}\cdot\frac{y^4}{y^{-2}}\cdot\frac{y^{12}}{y^7}=y^{3-6}\cdot y^{4-(-2)}\cdot y^{12-7}=y^{-3}\cdot y^6\cdot y^5$

When in the second stage we applied the aforementioned law of exponents for the second fraction (from left to right) carefully, this is because the term in the denominator of this fraction has a negative exponent and according to the aforementioned law of exponents, we need to subtract between the exponent of the numerator and the exponent of the denominator, which in this case gave us subtraction of a negative number from another number, an operation we performed carefully.

From here on we will no longer indicate the multiplication sign, but use the conventional writing form where placing terms next to each other means multiplication.

Let's return to the problem and note that we need to perform multiplication between terms with identical bases, therefore we will use the law of exponents for multiplication between terms with identical base:

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

Note that this law can only be used to calculate the multiplication being performed between terms with identical bases.

Let's apply this law in the problem:

$y^{-3}y^6y^5=y^{-3+6+5}=y^8$

We got the most simplified expression possible and therefore we are done,

Therefore the correct answer is B.

To solve this problem, we need to simplify and compare the given expressions.

Let's simplify each:

• $y^7 \times y^2$:
  Using the product of powers rule, $y^7 \times y^2 = y^{7+2} = y^9$.
• $(y^4)^3$:
  Using the power of a power rule, $(y^4)^3 = y^{4 \times 3} = y^{12}$.
• $y^9$:
  This is already in its simplest form, $y^9$.
• $\frac{y^{11}}{y^4}$:
  Using the power of a quotient rule, $\frac{y^{11}}{y^4} = y^{11-4} = y^7$.

Now that all the expressions are in the form $y^n$, we can compare the exponents to see which is greatest: $y^9$, $y^{12}$, $y^9$, and $y^7$.

The expression with the highest power is $y^{12}$, which corresponds to the choice $(y^4)^3$.

Thus, the greater value among the choices is $(y^4)^3$.

To determine which value is greater, let's simplify each choice:

Choice 1: $(a^2)^4$
By using the power of a power rule: $(x^m)^n = x^{m \times n}$, it simplifies to:
$(a^2)^4 = a^{2 \times 4} = a^8$.

Choice 2: $a^2 + a^0$
Evaluate using the zero exponent rule, $a^0 = 1$:
This expression becomes $a^2 + 1$.

Choice 3: $a^2 \times a^1$
Apply the product of powers rule: $x^m \times x^n = x^{m+n}$:
This simplifies to $a^{2+1} = a^3$.

Choice 4: $\frac{a^{14}}{a^9}$
Apply the quotient of powers rule: $\frac{x^m}{x^n} = x^{m-n}$:
This simplifies to $a^{14-9} = a^5$.

Now, let's compare these simplified forms:
We have $a^8$, $a^2 + 1$, $a^3$, and $a^5$.

For $a > 1$, exponential functions grow rapidly, thus:
- $a^8$ is greater than $a^5$.
- $a^8$ is greater than $a^3$.
- $a^8$ is greater than $a^2 + 1$ for sufficiently large $a$.

Thus, the expression with the highest power, and therefore the greatest value, is $(a^2)^4$.

Let's first deal with the first term in the multiplication, noting that the terms in the numerator and denominator have identical bases, so we'll use the power rule for division between terms with the same base:

$\frac{a^m}{a^n}=a^{m-n}$We'll apply for the first term in the expression:

$\frac{a^{3b}}{a^{2b}}\cdot a^b=a^{3b-2b}\cdot a^b=a^b\cdot a^b$where we also simplified the expression we got as a result of subtracting the exponents of the first term,

Next, we'll notice that the two terms in the multiplication have identical bases, so we'll use the power rule for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We'll apply to the problem:

$a^b\cdot a^b=a^{b+b}=a^{2b}$Therefore, the correct answer is A.

Write the problem in an organized way using fraction notation for the first term:$\frac{}{}\frac{X^3\cdot X^2}{X^5}+X^4$

Let's continue and refer to the first term in the above sum:

$\frac{X^3\cdot X^2}{X^5}$

Begin with the numerator, using the law of exponents for multiplying terms with identical bases:

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

and we obtain the following:

$\frac{X^3\cdot X^2}{X^5}=\frac{X^{3+2}}{X^5}=\frac{X^5}{X^5}$

Now proceed to use the law of exponents for the division between terms with identical bases:

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

When in the first stage of the above formula we just wrote the same thing in fraction notation instead of using division (:), let's apply the law of exponents to the problem and calculate the result for the first term that we obtained above:

$\frac{X^5}{X^5}=X^{5-5}=X^0$

Proceed to apply the law of exponents:

$a^0=1$

Note that this rule is actually just the understanding that dividing a number by itself will always give the result 1. Let's return to the problem and we obtain the result of the first term in the exercise (meaning - the result of calculating the fraction) is:

$X^0=1$,

Let's return to the complete exercise and summarize everything said so far as follows:

$\frac{X^3\cdot X^2}{X^5}+X^4=\frac{X^5}{X^5}+X^4=X^0+X^4=1+X^4$

First we rewrite the first expression on the left of the problem as a fraction:

$\frac{a^2}{a}+a^3\cdot a^5$Then we use two properties of exponentiation, to multiply and divide terms with identical bases:

1.

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

2.

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

Returning to the problem and applying the two properties of exponentiation mentioned earlier:

$\frac{a^2}{a}+a^3\cdot a^5=a^{2-1}+a^{3+5}=a^1+a^8=a+a^8$

Later on, keep in mind that we need to factor the expression we obtained in the last step by extracting the common factor,

Therefore, we extract from outside the parentheses the greatest common divisor to the two terms which are:

$a$ We obtain the expression:

$a+a^8=a(1+a^7)$ when we use the property of exponentiation mentioned earlier in A.

$$$a^8=a^{1+7}=a^1\cdot a^7=a\cdot a^7$

Summarizing the solution to the problem and all the steps, we obtained the following:

$\frac{a^2}{a}+a^3\cdot a^5=a(1+a^{7)}$

Therefore, the correct answer is option b.

We'll begin by applying the multiplication law between fractions, multiplying numerator by numerator and denominator by denominator:

$\frac{x}{y}\cdot\frac{w}{z}=\frac{x\cdot w}{y\cdot z}$

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

$\frac{a^{12}}{a^9}\cdot\frac{a^3}{a^4}=\frac{a^{12}\cdot a^3}{a^{^9}\cdot a^4}$

From here on we will no longer indicate the multiplication sign, instead we will place the terms next to each other.

Note that in both the numerator and denominator, multiplication is performed between terms with identical bases, therefore we'll apply the power law for multiplication between terms with the same base:

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

Note that this law can only be used to calculate multiplication between terms with identical bases.

Let's return to the problem and calculate separately the results of the multiplication in the numerator and denominator:

$\frac{a^{12}a^3}{a^{^9}a^4}=\frac{a^{12+3}}{a^{9+4}}=\frac{a^{15}}{a^{13}}$

In the last step we calculated the sum of the exponents.

Now we need to perform division (fraction=division operation between numerator and denominator) between terms with identical bases, therefore we'll apply the power law for division between terms with the same base:

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

Note that this law can only be used to calculate division between terms with identical bases.

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

$\frac{a^{15}}{a^{13}}=a^{15-13}=a^2$

In the last step we calculated the result of the subtraction operation in the exponent.

We cannot simplify the expression further. Therefore the correct answer is D.

To solve this problem, we need to simplify the given expression using the rules of exponents:

First, simplify inside the brackets:
$\frac{a^4}{a^3} \times \frac{a^8}{a^7} = a^{4-3} \times a^{8-7} = a^1 \times a^1 = a^{1+1} = a^2$

Now, handle the entire expression, dividing it by $\frac{a^{10}}{a^8}$:
$\frac{a^2}{\frac{a^{10}}{a^8}} = a^2 \times \frac{a^8}{a^{10}} = a^2 \times a^{8-10} = a^2 \times a^{-2} = a^{2 + (-2)} = a^0$

Recall that any non-zero number raised to the power of zero is 1, hence: $a^0 = 1$

Therefore, the solution to the problem is $1$.

Let's start with multiplying the fractions, remembering that the multiplication of fractions is performed by multiplying the numerator by numerator and the denominator by the denominator:

$\frac{b^{22}}{b^{20}}\cdot\frac{b^{30}}{b^{20}}=\frac{b^{22}\cdot b^{30}}{b^{20}\cdot b^{20}}$

In both the numerator and denominator, multiplication occurs between terms with identical bases, so we'll apply the power law for multiplying terms with identical bases:

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

This law can only be used when multiplication is performed between terms with identical bases.

From here on, we will no longer indicate the multiplication sign, instead we will place terms next to each other.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

$\frac{b^{22}b^{30}}{b^{20}b^{20}}=\frac{b^{22+30}}{b^{20+20}}=\frac{b^{52}}{b^{40}}$

In the final step we calculated the sum of the exponents in the numerator and denominator.

Note that division is required between two terms with identical bases, hence we'll apply the power law for division between terms with identical bases:

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

This law can only be used when division is performed between terms with identical bases.

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

$\frac{b^{52}}{b^{40}}=b^{52-40}=b^{12}$

In the final step we calculated the subtraction between the exponents.

This is the most simplified form of the expression:

Therefore, the correct answer is C.

Let's start with multiplying the fractions, remembering that the multiplication of fractions is performed by multiplying the numerator by the numerator and the denominator by the denominator:

$\frac{a^{20b}}{a^{15b}}\cdot\frac{a^{3b}}{a^{2b}}=\frac{a^{20b}\cdot a^{3b}}{a^{15b}\cdot a^{2b}}$

In both the numerator and denominator, multiplication occurs between terms with identical bases, thus we'll apply the power law for multiplying terms with identical bases:

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

We emphasize that this law can only be used when multiplication is performed between terms with identical bases.

From this point forward, we will no longer use the multiplication sign, instead we will place terms next to each other.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

$\frac{a^{20b}a^{3b}}{a^{15b}a^{2b}}=\frac{a^{20b+3b}}{a^{15b+2b}}=\frac{a^{23b}}{a^{17b}}$

In the final step we calculated the sum of the exponents in the numerator and denominator.

Now we need to perform division between two terms with identical bases, thus we'll apply the power law for dividing terms with identical bases:

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

This law can only be used when division is performed between terms with identical bases.

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

$\frac{a^{23b}}{a^{17b}}=a^{23b-17b}=a^{6b}$

In the final step we calculate the subtraction between the exponents.

This is the most simplified form of the expression:

Therefore, the correct answer is D.

Apply the power law for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We'll apply this law to the problem:

$\big(-\frac{1}{8}\big)^8\cdot\big(-\frac{1}{8}\big)^{-3}=\big(-\frac{1}{8}\big)^{8+(-3)}=\big(-\frac{1}{8}\big)^{8-3}=\big(-\frac{1}{8}\big)^5$

In the first stage we applied the above power law and in the following stages we simplified the expression in the exponent,

Let's continue and use the power law for power of terms in parentheses:

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

We'll apply this law to the expression that we obtained in the last stage:

$\big(-\frac{1}{8}\big)^5=\big(-1\cdot\frac{1}{8}\big)^5=(-1)^5\cdot\big(\frac{1}{8}\big)^5=-1\cdot\big(\frac{1}{8}\big)^5=-\big(\frac{1}{8}\big)^5$

In the first stage we presented the expression in parentheses as a multiplication between negative one and a positive number. In the next stage we applied the above power law and then simplified the expression we obtained whilst noting that negative one to an odd power will (always) give the result negative one.

Next we'll recall two additional power laws:

a. The negative power law:

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

b. The power law for power of a power:

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

We'll continue and apply these two laws to the expression that we obtained in the last stage:

$-\big(\frac{1}{8}\big)^5=-(8^{-1})^5=-8^{(-1)\cdot5}=-8^{-5}$

In the first stage we presented the fraction inside the parentheses as a term with a negative power using the above power law for negative power mentioned in a. above. In the next stage we applied the power law for power of a power mentioned in b. above carefully, given that the term inside the parentheses has a negative power. We then simplified the expression in the exponent.

Let's summarize the solution :

$\big(-\frac{1}{8}\big)^8\cdot\big(-\frac{1}{8}\big)^{-3}=\big(-\frac{1}{8}\big)^5=-\big(\frac{1}{8}\big)^5=-(8^{-1})^5=-8^{-5}$

Therefore the correct answer is answer d.

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

$b^m\cdot b^n=b^{m+n}$We'll apply this law to the fraction in the expression in the problem:

$\frac{a^4a^8a^{-7}}{a^9}=\frac{a^{4+8+(-7)}}{a^{^9}}=\frac{a^{4+8-7}}{a^9}=\frac{a^5}{a^9}$where in the first stage we'll apply the aforementioned law of exponents and in the following stages we'll simplify the resulting expression,

Let's now recall the law of exponents for division between terms with identical bases:

$\frac{b^m}{b^n}=b^{m-n}$We'll apply this law to the expression we got in the last stage:

$\frac{a^5}{a^9}=a^{5-9}=a^{-4}$Let's now recall the law of exponents for negative exponents:

$b^{-n}=\frac{1}{b^n}$And we'll apply this law of exponents to the expression we got in the last stage:

$a^{-4}=\frac{1}{a^4}$Let's summarize the solution steps so far, we got that:

$\frac{a^4a^8a^{-7}}{a^9}=\frac{a^5}{a^9}=\frac{1}{a^4}$Therefore, the correct answer is answer A.

Let's deal with the first term in the problem, which is the fraction,

For this, we'll recall two laws of exponents:

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

$a^m\cdot a^n=a^{m+n}$b. The law of exponents for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$Let's apply these laws of exponents to the problem:

$$$\frac{17^{-3}\cdot17^{3x}}{17}-17x=\frac{17^{-3+3x}}{17}-17x=17^{-3+3x-1}-17x=17^{3x-4}-17x$where in the first stage we'll apply the law of exponents mentioned in 'a' above to the fraction's numerator, and in the next stage we'll apply the law of exponents mentioned in 'b' to the resulting expression, then we'll simplify the expression.

Therefore, the correct answer is answer a.

To determine which of the given expressions is the greatest, we will use the relevant exponent rules to simplify each one:

• Simplify $y^7 \times y^2$:
  Using the Product of Powers rule, we have $y^7 \times y^2 = y^{7+2} = y^9$.
• Simplify $(y^4)^3$:
  Using the Power of a Power rule, we have $(y^4)^3 = y^{4 \times 3} = y^{12}$.
• Simplify $y^9$:
  This expression is already simplified and is $y^9$.
• Simplify $\frac{y^{11}}{y^4}$:
  Using the Division of Powers rule, we have $\frac{y^{11}}{y^4} = y^{11-4} = y^7$.

After simplifying, we compare the powers of $y$ from each expression:

• $y^9$ from $y^7 \times y^2$
• $y^{12}$ from $(y^4)^3$
• $y^9$ from $y^9$
• $y^7$ from $\frac{y^{11}}{y^4}$

Clearly, $y^{12}$ is the largest power among the expressions, meaning that $(y^4)^3$ is the greatest value.

Therefore, the correct choice is $(y^4)^3$.

To determine which expression has the greatest value, we apply the exponent rules to simplify each choice:

• For $x^3 \times x^4$, using the product rule: $x^3 \times x^4 = x^{3+4} = x^7$.
• For $(x^3)^5$, using the power of a power rule: $(x^3)^5 = x^{3 \times 5} = x^{15}$.
• $x^{10}$ is already in its simplest form.
• For $\frac{x^9}{x^2}$, using the quotient rule: $\frac{x^9}{x^2} = x^{9-2} = x^7$.

To identify the greater value, we compare the exponents:

• $x^7$ from choices 1 and 4.
• $x^{15}$ from choice 2.
• $x^{10}$ from choice 3.

The expression with the largest exponent is $(x^3)^5$ or $x^{15}$.

Therefore, the expression with the greatest value is $(x^3)^5$.

First, simplify the numerator and the denominator separately:
Numerator: $X^4 \cdot X^3 = X^{4+3} = X^7$
Denominator: $X^5 \cdot X^2 = X^{5+2} = X^7$

Now, combine the simplified numerator and denominator:

$\frac{X^7}{X^7}$

Since any number divided by itself is 1, we have:

$\frac{X^7}{X^7} = 1$

Therefore, the correct answer is:

$1$

We use the three appropriate power properties to solve the problem:

1. Power law for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$ 2. Power law for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$ 3. Power law for the division of terms with identical bases:

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

We continue and apply the three previous laws to the problem:

$2^3\cdot2^4+(4^3)^2+\frac{2^5}{2^3}=2^{3+4}+4^{3\cdot2}+2^{5-3}=2^7+4^6+2^2$

In the first step we apply the power law mentioned in point 1 to the first expression on the left, the power law mentioned in point 2 to the second expression on the left, and the power law mentioned in point 3 to the third expression on the left, separately. In the second step, we simplify the expressions by exponents possession of the received terms,

Then,after using the substitution property for addition, we find that the correct answer is D.

Let's start with multiplying the two fractions in the problem using the rule for fraction multiplication, which states that we multiply numerator by numerator and denominator by denominator while keeping the fraction line:

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

Let's apply this rule to the problem and perform the multiplication between the fractions:

$\frac{16x^4}{5y}\cdot\frac{10y^2}{3x^4y}=\frac{16\cdot10\cdot x^4y^2}{5\cdot3\cdot x^4y\cdot y}=\frac{160x^4y^2}{15x^4y^2}$

Where in the first stage we performed the multiplication between the fractions using the above rule and then simplified the expressions in the numerator and denominator of the resulting fraction using the distributive property of multiplication and the law of exponents for multiplying terms with identical bases:

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

Which we applied in the last stage to the denominator of the resulting fraction.

Now we'll use the same rule for fraction multiplication again, but in the opposite direction in order to express the resulting fraction as a multiplication of fractions where each fraction contains only numbers or terms with identical bases:

$\frac{160x^4y^2}{15x^4y^2}=\frac{160}{15}\cdot\frac{x^4}{x^4}\cdot\frac{y^2}{y^2}$

We did this so we could continue and simplify the expression using 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 last expression we got:

$\frac{160}{15}\cdot\frac{x^4}{x^4}\cdot\frac{y^2}{y^2}=\frac{32}{3}\cdot x^{4-4}\cdot y^{2-2}=\frac{32}{3}x^0y^0$

Where in the first stage, in addition to applying the above law of exponents, we also simplified the numerical fraction after identifying that both its numerator and denominator are multiples of 5, and then simplified the resulting expression,

In the next stage we'll recall that raising any number to the power of 0 gives the result 1, meaning mathematically that:

$X^0=1$

Let's return to the expression we got and continue simplifying using this fact:

$\frac{32}{3}x^0y^0=\frac{32}{3}\cdot1\cdot1=\frac{32}{3}$

We can now convert the improper fraction we got to a mixed number to get:

$\frac{32}{3}=10\frac{2}{3}$

Let's summarize the solution to the problem, we got that:

$\frac{16x^4}{5y}\cdot\frac{10y^2}{3x^4y}=\frac{160x^4y^2}{15x^4y^2}=10\frac{2}{3}$

Therefore the correct answer is answer C.

Important Note:

In solving the problem above, we detailed the steps to the solution, and used fraction multiplication in both directions multiple times and the above law of exponents,

We could have shortened the process, applied the distributive property of multiplication, and performed directly both the application of the above law of exponents and the simplification of the numerical part to get directly the last line we got:

$\frac{16x^4}{5y}\cdot\frac{10y^2}{3x^4y}=10\frac{2}{3}$

(Meaning we could have skipped the part where we expressed the fraction as a multiplication of fractions and even the initial fraction multiplication we performed and immediately simplified between the fractions)

However, it should be emphasized that this quick solution method is conditional on the fact that between all terms in the numerator and denominator of each fraction in the problem, and also between the fractions themselves, multiplication is performed, meaning that we can put a single fraction line like we did at the beginning and can apply the distributive property of multiplication and so on, this is a point worth noting, since not every problem we encounter will meet all the conditions mentioned here in this note.

To solve this problem, follow these steps:

Step 1: Simplify the first fraction:

The first expression is $\frac{35x \cdot y^7}{7xy}$.

• Cancel the common factor of $7$: $35 \div 7 = 5$.

• This simplifies to $\frac{5x \cdot y^7}{x \cdot y}$.

• Cancel the common factor of $x$: $x/x$ cancels to $1$.

• Cancel part of the $y$ terms: $y^7/y = y^{7-1} = y^6$.

• The result is $5y^6$.

Step 2: Simplify the second fraction:

The second expression is $\frac{8x}{5y}$.

• No common factors in the numerator and denominator, so it remains $\frac{8x}{5y}$.

Step 3: Multiply these simplified results:

Now, multiply the results from Step 1 and Step 2: $5y^6 \cdot \frac{8x}{5y}$.

• The factor of $5$ in $5y^6$ and $\frac{8x}{5y}$ cancels: $5/5 = 1$.

• This results in $y^6 \cdot \frac{8x}{y}$.

• Cancel part of the $y$ terms: $y^6/y = y^{6-1} = y^5$.

Thus, the simplified expression is $8xy^5$.

Therefore, the solution to the problem is $\mathbf{8xy^5}$.

Let's start with multiplying the two fractions in the problem using the rule for fraction multiplication, which states that we multiply numerator by numerator and denominator by denominator while keeping the fraction line:

$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$Let's apply this rule to our problem and perform the multiplication between the fractions:

$\frac{38\cdot x^5y^4}{9x}\cdot\frac{5xy}{3y^2}=\frac{38\cdot5\cdot x^5xy^4y}{9\cdot3\cdot xy^2}=\frac{190\cdot x^6y^5}{27\cdot xy^2}$

In the first stage, we performed the multiplication between the fractions using the above rule, and then simplified the expressions in the numerator and denominator of the resulting fraction by using the distributive property of multiplication and the law of exponents for multiplying terms with identical bases:

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

We applied this in the final stage to the fraction's numerator.

Now we'll use the same rule for fraction multiplication again, but in the opposite direction, in order to express the resulting fraction as a multiplication of fractions where each fraction contains only numbers or terms with identical bases:

$\frac{190\cdot x^6y^5}{27\cdot xy^2}=\frac{190}{27}\cdot\frac{x^6}{x}\cdot\frac{y^5}{y^2}$

We did this so we could continue and simplify the expression using the law of exponents for division between terms with identical bases:

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

Let's apply the above law to the last expression we got:

$\frac{190}{27}\cdot\frac{x^6}{x}\cdot\frac{y^5}{y^2}=\frac{190}{27}x^{6-1}y^{5-2}=\frac{190}{27}\cdot x^5y^3=7\frac{1}{27}\cdot x^5y^3$

In the first stage we applied the above law of exponents, then simplified the resulting expression, additionally we removed the multiplication sign and switched to the conventional multiplication notation by placing the terms next to each other, and in the final stage we converted the improper fraction we got at the beginning of the last expression to a mixed number.

Let's summarize the solution to the problem, we got that:

$\frac{38\cdot x^5y^4}{9x}\cdot\frac{5xy}{3y^2}= \frac{190\cdot x^6y^5}{27\cdot xy^2} =7\frac{1}{27}\cdot x^5y^3$

Therefore the correct answer is answer B.

Important note:

In solving the problem above, we detailed the steps to the solution, and used fraction multiplication in both directions and multiple times along with the mentioned law of exponents,

We could have shortened the process, applied the distributive property of multiplication, and performed directly both the application of the mentioned law of exponents and the numerical part reduction to get directly the last line we received:

$\frac{38\cdot x^5y^4}{9x}\cdot\frac{5xy}{3y^2}=7\frac{1}{27}\cdot x^5y^3$

(Meaning we could have skipped the part where we expressed the fraction as a multiplication of fractions and even the initial fraction multiplication we performed and immediately perform the reduction between the fractions)

However, it should be emphasized that this quick solution method is conditional on the fact that between all terms in the numerator and denominator of each fraction in the problem, and also between the fractions themselves, multiplication is performed, meaning that we can enter a single fraction line as we did at the beginning and can apply the distributive property and express as fraction multiplication etc., this is a point worth noting, since not every problem we encounter will meet all the conditions mentioned here in this note.