Multiplication of Powers: Two Variables

To solve this problem, we'll follow these steps:

• Step 1: Identify the exponents in the given expression.
• Step 2: Use the property of exponents for multiplication by like bases.
• Step 3: Add the exponents together and simplify.

Now, let's work through each step:

Step 1: The original expression is $10^{a+b} \times 10^{a+1} \times 10^{b+1}$.

Step 2: Since the base (10) is the same for all terms, we add the exponents:

Step 3: Simplifying further:

Thus, the expression simplifies to:

Therefore, the solution to the problem is $10^{2b+2a+2}$.

To solve this problem, we'll follow these steps:

• Step 1: Confirm the given expression is $4^x \times 4^2 \times 4^a$.
• Step 2: Apply the exponent rule for multiplication of powers: if $b^m \times b^n = b^{m+n}$, use this with base 4.
• Step 3: Add the exponents of each term.

Let's work through these steps:

Step 1: The expression we have is $4^x \times 4^2 \times 4^a$.

Step 2: Since all parts of the product have the same base $4$, we can use the rule for multiplying powers: $4^x \times 4^2 \times 4^a = 4^{x+2+a}$.

Step 3: The simplified expression is obtained by adding the exponents: $x + 2 + a$.

Therefore, the expression $4^x \times 4^2 \times 4^a$ simplifies to $4^{x+2+a}$.

To solve this problem, we'll start by applying the rules for multiplying powers with the same base.

• Step 1: Identify the expression given as $7^{x+a} \times 7^a \times 7^x$.
• Step 2: Apply the rule $a^m \times a^n = a^{m+n}$ to combine the exponents since all terms have the same base, 7.

Let's proceed to simplify:

Combine the exponents:

$7^{x+a} \times 7^a \times 7^x = 7^{(x+a) + a + x}$.

Now, simplify the addition of the exponents:

$7^{(x+a) + a + x} = 7^{x + a + a + x}$.

Combine like terms in the exponent:

$x + a + a + x = 2x + 2a$.

Thus, the expression simplifies to:

$7^{2x+2a}$.

Therefore, the simplification of the given expression is $7^{2x+2a}$.

To solve this problem, let's simplify the expression $8^{3x} \times 8^{3y} \times 8^{2y+x}$ using exponent rules:

Step 1: Identify the exponents in each term:
- The first term is $8^{3x}$ with an exponent of $3x$.
- The second term is $8^{3y}$ with an exponent of $3y$.
- The third term is $8^{2y+x}$ with an exponent of $2y+x$.

Step 2: Apply the multiplication of powers rule:
Since all terms have the same base of 8, add the exponents: $(3x) + (3y) + (2y + x)$.

Step 3: Simplify the expression:
Adding the terms in the exponent gives us: $3x + 3y + 2y + x = 4x + 5y$.

Therefore, the simplified expression is $8^{4x+5y}$.

To solve this problem, we'll use the property of exponents for multiplying powers with the same base:

• Step 1: Identify that all terms have the same base, which is $8$. The equation is given as $8^a \times 8^2 \times 8^x$.

• Step 2: Apply the multiplication property of exponents: $b^m \times b^n = b^{m+n}$.

• Step 3: Add the exponents: $(a) + (2) + (x)$ to get the new exponent for the single base.

By applying these steps, we obtain:

$8^{a+2+x}$

This result matches choice 1, confirming that this is the correct simplified expression.

To solve this problem, we'll follow these steps:

• Step 1: Recognize that all the terms have the same base, which is 9.
• Step 2: Apply the exponent addition rule to combine the exponents.
• Step 3: Simplify the result by performing algebraic addition on the exponents.

Now, let's work through each step:
Step 1: The problem gives us the expression $9^{2a} \times 9^{2x} \times 9^a$. All terms share the same base, which is 9.
Step 2: Using the property of exponents $b^m \times b^n = b^{m+n}$, add the exponents: $(2a) + (2x) + a$.
Step 3: Combine like terms: $2a + a + 2x = 3a + 2x$. So, the expression becomes $9^{3a+2x}$.

Therefore, the simplified form of the given equation is $9^{3a+2x}$.

In the exercise of multiplying powers, we will add up all the powers of the same product, in this case the terms a, b

We use the formula:

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

We are going to focus on the term a:

$a^3\times a^2=a^{3+2}=a^5$

We are going to focus on the term b:

$b^4\times b^5=b^{4+5}=b^9$

Therefore, the exercise that will be obtained after simplification is:

$a^5\times b^9$

First, we'll use the distributive property of multiplication and arrange the algebraic expression according to like bases:

$a^4a^5b^5$

Next, we'll use the laws of exponents to multiply terms with like bases:

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

Therefore, we can combine all terms with the same base under one base:

$a^{4+5}b^5=a^9b^5$

Note that we could only combine terms with identical bases using this law,

From here we can observe that the expression cannot be simplified further, and therefore this is the correct answer, which is answer B (since the distributive property of multiplication is satisfied).

Important Note:

Note that for multiplication between numerical terms, we can denote the multiplication operation using a dot ($\cdot$), known as dot-product, or using the "times" symbol ($\times$) known as cross-product. For numerical terms, these operations are identical. We can also indicate multiplication by placing the terms next to each other without explicitly writing the operation between them. In such cases, there is a universal understanding that this represents multiplication between the terms. Usually, the multiplication is not explicitly noted (meaning the last option we mentioned here), and if it is noted, dot notation is typically used. In this problem, both in the question and answer, they chose to use cross notation, but the meaning is always the same since we are dealing with numerical terms.

Begin by applying the distributive property of multiplication and proceed to arrange the algebraic expression according to like bases:

$a^ya^xa^{6\text{ }}b^97^y$

Next, we'll use the power rule to multiply terms with the same base:

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

Note that this rule applies to any number of terms in multiplication, not just two. For example, when multiplying three terms with the same base, we obtain the following:

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

Therefore, we can combine all terms with the same base under one base:

$a^{y+x+6}b^97^y$

Note that we could only combine terms with identical bases using this rule,

From here we can see that the expression cannot be simplified further, and therefore this is the correct answer, which is answer C (since the distributive property of multiplication holds).

Using the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It is important to note that this law is only valid for terms with identical bases,

We notice that in the problem there are two types of terms. First, for the sake of order, we will use the substitution property to rearrange the expression so that the two terms with the same base are grouped together. The, we will proceed to solve:

$k^2t^4k^6t^2=k^2k^6t^4t^2$Next, we apply the power property to each different type of term separately,

$k^2k^6t^4t^2=k^{2+6}t^{4+2}=k^8t^6$We apply the property separately - for the terms whose bases are$k$and for the terms whose bases are$t$We add the powers in the exponent when we multiply all the terms with the same base.

The correct answer then is option b.

To simplify the given mathematical expression, we'll follow these steps:

• Step 1: Apply the Product of Powers Property to terms with the same base. For the expression $a^{-5} \times a^8$, we use the rule:
• $a^m \times a^n = a^{m+n}$.
• Step 2: Add the exponents: $-5 + 8 = 3$. Therefore, $a^{-5} \times a^8 = a^3$.
• Step 3: Since $x^3$ does not share the base $a$, it remains as is in the expression.

Therefore, the simplified form of the expression $a^{-5} \times a^8 \times x^3$ is:

$a^3 \times x^3$

To reduce the given equation $a^{-3x} \times a^b \times a^b$, we will use the multiplication of powers rule for exponents, which states that if you multiply powers with the same base, you add the exponents.

Let's follow the steps:

• Step 1: Identify that all terms share the same base, $a$.

• Step 2: Apply the rule: $a^{-3x} \times a^b \times a^b = a^{-3x + b + b}$.

• Step 3: Simplify the exponents by adding them: $-3x + b + b = -3x + 2b$.

Therefore, the reduced form of the equation is $a^{-3x + 2b}$.

$$We will now apply the law of negative exponents, but in reverse:

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

We'll apply this law to the problem for the third term in the product:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n}\cdot n^{-m}\cdot m^{-1}=m^{-n}\cdot m^{-1}\cdot n^{-m}$

When in the first stage we applied the above law of exponents for the third term in the product, and in the next stage we rearranged the resulting expression using the distributive property of multiplication so that terms with identical bases are adjacent to each other,

Next, we'll recall the law of exponents for multiplying terms with identical bases:

$a^x\cdot a^y=a^{x+y}$

And we'll apply this law of exponents to the expression that we obtained in the last stage:

$m^{-n}\cdot m^{-1}\cdot n^{-m}=m^{-n+(-1)}\cdot n^{-m}=m^{-n-1}\cdot n^{-m}$

When in the first stage we applied the above law of exponents for terms with identical bases, and in the next stage we simplified the expression with the exponent of the first term in the product in the expression we obtained the following,

Let's summarize the solution so far as shown below:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n}\cdot n^{-m}\cdot m^{-1}=m^{-n-1}\cdot n^{-m}$

Now let's note that there is no such answer among the given options, and an additional 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 given options,

Let's note that options B and D have expressions similar to the expression we got in the last stage, while the other two options can be directly eliminated since they are clearly different from the expression we got,

Furthermore, let's note that in addition, the second term in the product in the expression we got, which is the term-$n^{-m}$, is in the numerator (note at the end of the solution on this topic), while in option B it's in the denominator, so we'll eliminate this option,

Thus - we're left with only one option - which is answer D, however we want to verify (and must verify!) that this is indeed the correct answer:

We'll do this using the law of exponents for negative exponents that we mentioned earlier, but in the forward direction:

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

And we'll deal separately with the first term in the product in the expression we got in the last stage of solving the problem, which is the term:

$m^{-n-1}$

Let's note that we can represent the expression in the exponent as follows:

$-n-1=-(n+1)$

Where we used factoring out and took out negative one from the parentheses,

Next, we'll use the above law of exponents and the last understanding to represent the above expression (which we're currently dealing with, separately) as a term in the denominator of a fraction:

$m^{-n-1}=m^{-\underline {\bm{(n+1)}}}=\frac{1}{m^{\underline {\bm{n+1}}}}$

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

$-n-1=-(n+1)$,

Next, we applied the above law of exponents carefully, since the number that- x represents in our use of the above law of exponents here is:

$n+1$(underlined in the expression above)

Let's return then to the expression we got in the last stage of solving the given problem, and apply for the first term in the product the mathematical manipulation we just performed:

$m^{-n-1}\cdot n^{-m}=m^{-(n+1)}\cdot n^{-m}=\frac{1}{m^{n+1}}\cdot n^{-m}$

Now let's simplify the expression that we obtained and perform the multiplication in the fraction while remembering that multiplication in a fraction means multiplying the numerators:

$\frac{1}{m^{n+1}}\cdot n^{-m}=\frac{1\cdot n^{-m}}{m^{n+1}}=\frac{n^{-m}}{m^{n+1}}$

Let's summarize then the solution stages so far as follows:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n-1}\cdot n^{-m} =\frac{n^{-m}}{m^{n+1}}$

Therefore, the correct answer is indeed answer D.

Note:

When it's written "the number in the numerator" even though there isn't actually a fraction in the expression, this is because we can always refer to any number as being in the numerator of a fraction if we remember that any number divided by 1 equals itself, meaning, we can always write a number as a fraction like this:

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

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.

Begin by calculating the result of the multiplication inside of the parentheses in the first term of the multiplication and proceed to write down the entire expression:

$(5\cdot6\cdot4)^x\cdot3^x\cdot4^y=120^x\cdot3^x\cdot4^y$
Note that the first two terms in the multiplication have the same exponent, hence we can apply the law of exponents for parentheses, however in the opposite direction:

$z^n\cdot t^n=(z\cdot t)^n$
This means that instead of opening the parentheses whilst applying the (same) exponent to each term of the multiplication inside the parentheses. We'll place the two terms (with identical exponents) as a multiplication inside of the parentheses under the exponent. This is possible in this problem given that the first two terms in the multiplication have identical exponents,

Let's apply it to the problem:

$120^x\cdot3^x\cdot4^y =(120\cdot3)^x\cdot4^y$
Now we can of course calculate the result of the multiplication inside the parentheses if we want, and simplify the resulting expression even further, but at this stage it's worth noting that this result is answer A, and there is no other answer among the options that is both correct and more simplified,

Therefore, the correct answer is A.

To solve this problem, we'll proceed with the following steps:

• Step 1: Simplify each fraction separately.

Consider the first fraction:

$\frac{15x^4y^3}{8x^2y^5}$

Apply the quotient rule of exponents: $\frac{x^m}{x^n} = x^{m-n}$ and $\frac{y^m}{y^n} = y^{m-n}$.

This gives us: $\frac{15}{8} \cdot x^{4-2} \cdot y^{3-5} = \frac{15}{8} \cdot x^2 \cdot y^{-2}$.

• Step 2: Simplify the second fraction.

Consider the second fraction:

$\frac{24yx^7}{3xy^2}$

Apply the quotient rule: $\frac{24}{3} \cdot \frac{y}{y^2} \cdot \frac{x^7}{x^1} = 8 \cdot y^{1-2} \cdot x^{7-1} = 8 \cdot y^{-1} \cdot x^6$.

• Step 3: Multiply the simplified fractions together.

Now, multiply the results:

$\left(\frac{15}{8} \cdot x^2 \cdot y^{-2}\right) \cdot \left(8 \cdot y^{-1} \cdot x^6\right)$

Simplify by multiplying coefficients and applying exponent rules: $\frac{15 \times 8}{8} \cdot x^{2+6} \cdot y^{-2-1}$.

Which simplifes to: $15 \cdot x^8 \cdot y^{-3}$.

Therefore, the expression simplifies to $15x^8y^{-3}$.

Finally, matching this result with the provided choices, we find that the correct answer is choice (3):

$15x^8y^{-3}$

To solve this problem, we'll follow these steps:

• Step 1: Simplify each fraction separately.
• Step 2: Multiply the simplified fractions together.
• Step 3: Cancel common terms if necessary.
• Step 4: Apply exponent rules for a clearer expression.

Step 1: Simplify each fraction:

The first fraction is $\frac{27yx}{3x^2}$. This can be simplified as follows:

$\frac{27yx}{3x^2} = \frac{27}{3} \cdot \frac{yx}{x^2} = 9 \cdot \frac{y}{x}$.

The second fraction is $\frac{5y^4x^2}{3y}$. Simplifying it, we have:

$\frac{5y^4x^2}{3y} = \frac{5x^2}{3} \cdot y^{4-1} = \frac{5x^2}{3} \cdot y^3$.

Step 2: Multiply the simplified fractions:

$9 \cdot \frac{y}{x} \times \frac{5x^2}{3} \cdot y^3 = \frac{9 \cdot 5x^2 \cdot y \cdot y^3}{3 \cdot x}$.

Step 3: Simplify again by cancelling out common terms:

$= \frac{9 \times 5 \cdot x \cdot y^{1+3}}{3} = \frac{45xy^4}{3}$.

Divide 45 by 3: $= 15xy^4$.

Therefore, the product of the two expressions simplifies to $15y^4x$, which matches choice 1.

To solve the problem, we'll follow these steps:

• Step 1: Multiply the fractions $\frac{78xy^5}{3x^5}$ and $\frac{4yx}{5y^4}$.

• Step 2: Simplify the coefficients and apply exponent rules to the variables.

• Step 3: Identify the correct multiple-choice option matching the simplified expression.

Now, let's work through each step in detail:

Step 1: Multiply the Fractions

We multiply the numerators together and the denominators together:

$\frac{78xy^5}{3x^5} \cdot \frac{4yx}{5y^4} = \frac{78xy^5 \cdot 4yx}{3x^5 \cdot 5y^4}$

Simplifying, we have:

$= \frac{78 \cdot 4 \cdot x \cdot x \cdot y^5 \cdot y}{3 \cdot 5 \cdot x^5 \cdot y^4}$

Step 2: Simplify the Expression

Simplify the coefficients:

$78 \cdot 4 = 312$ and $3 \cdot 5 = 15$

Combine the coefficients:

$\frac{312}{15}$

Now simplify the variables using exponent rules:

Combine powers of the same base:

$x \cdot x = x^2$

The numerator becomes:

$312 \cdot x^2 \cdot y^6$

The denominator according to $\frac{x^2}{x^n} = x^{2-n}$, given,

$15 \cdot x^5 \cdot y^4$

Simplifying the exponents:

$x^{2-5} = x^{-3}$ and $y^{6-4} = y^2$

Thus,

$\frac{312 \cdot x^{-3} \cdot y^2}{15}$

Conclusion:

After simplifying the expression, the result is:

$\frac{312\cdot x^{-3}\cdot y^2}{15}$

Matching this with the multiple-choice options, the correct choice is option 3.

Therefore, the solution to the problem is $\frac{312\cdot x^{-3}\cdot y^2}{15}$.

To solve this problem, we'll simplify each part of the given expression step by step:

Original expression:
$\frac{85x \cdot y^3}{5y^4x^3} \cdot \frac{9xy}{3yx^2}$.

Let's simplify the first fraction:
$\frac{85x \cdot y^3}{5y^4x^3}$.

• Divide the coefficients: $\frac{85}{5} = 17$.
• Apply the quotient rule of exponents:
  $\frac{x}{x^3} = x^{1-3} = x^{-2}$ and $\frac{y^3}{y^4} = y^{3-4} = y^{-1}$.
• This simplifies to: $17x^{-2}y^{-1}$.

Now, simplify the second fraction:
$\frac{9xy}{3yx^2}$.

• Divide the coefficients: $\frac{9}{3} = 3$.
• Cancel the $y$: $\frac{y}{y} = 1$.
• Apply the quotient rule of exponents: $\frac{x}{x^2} = x^{1-2} = x^{-1}$.
• This simplifies to: $3x^{-1}$.

Multiply the simplified expressions:
$(17x^{-2}y^{-1})\cdot (3x^{-1})$.

• Combine the coefficients: $17 \cdot 3 = 51$.
• Apply the product rule for $x$: $x^{-2} \cdot x^{-1} = x^{-2-1} = x^{-3}$.
• For $y$, simply note: $y^{-1}$.

Thus, the simplified expression is: $51x^{-3}y^{-1}$.

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.