Multiplication of Powers: Numbers as coefficients

Note that there is multiplication operation between all terms in the expression, hence we'll first apply the distributive property of multiplication in order to handle the coefficients of terms raised to powers, and the terms themselves separately. For greater clarity, let's break this down into steps:

$a^x\cdot3a^y\cdot a^2\cdot2a=3\cdot2\cdot a^x\cdot a^y\cdot a^2\cdot a=6\cdot a^x\cdot a^y\cdot a^2\cdot a$

Due to the multiplication operation between all terms we could do this, it should be noted that we can (and it's preferable to) skip the middle step, meaning:

Write directly:$a^x\cdot3a^y\cdot a^2\cdot2a=6\cdot a^x\cdot a^y\cdot a^2\cdot a$

From here on we won't write the multiplication sign anymore instead we simply place the terms next to each other\ place the term next to its coefficient to indicate multiplication between them,

Proceed to apply the law of exponents for multiplication of terms with identical bases:

$c^m\cdot c^n=c^{m+n}$Note also that this law applies to any number of terms being multiplied and not just two, for example for three terms with identical bases we obtain:

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

Whilst we used law of exponents twice, we can also perform the same calculation for four terms or 5 and so on..,

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

$6a^xa^yaa^2=6a^{x+y+2+1}=6a^{x+y+3}$

Therefore the correct answer is d.

Important note:

Here we need to emphasize that we should always ask the question - what is the exponent being applied to?

For example, in this problem the exponent applies only to the bases of-

$a$$$and not to the numbers, more clearly, in the following expression: $5c^7$the exponent applies only to $c$ and not to the number 5,

whereas when we write:$(5c)^7$the exponent applies to each term of the multiplication inside the parentheses,

meaning:$(5c)^7=5^7c^7$

This is actually the application of the law of exponents:

$(w\cdot r)^n=w^n\cdot r^n$

which follows both from the meaning of parentheses and from the definition of exponents.

Note that there is multiplication between all terms in the expression. Thus we'll first apply the distributive property of multiplication to understand that we can separately handle the coefficients of the terms raised to powers, and the terms themselves separately. For clarity, let's handle it in steps:

$b^3\cdot3b^2\cdot2b^{-2}=3\cdot2\cdot b^3\cdot b^2\cdot b^{-2}=6\cdot b^3\cdot b^2\cdot b^{-2}$

Given that there is multiplication between all terms, we could do this. It should be noted that we can (and it's preferable to) skip the middle step, meaning:

Write directly:$b^3\cdot3b^2\cdot2b^{-2}=6\cdot b^3\cdot b^2\cdot b^{-2}$

From here on we will no longer write the multiplication sign and remember that it is conventional to simply place the terms next to each other\ place the term next to its coefficient to indicate multiplication between them,

Proceed to apply the law of exponents for multiplication of terms with identical bases:

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

Note that this law applies to any number of terms being multiplied and not just two, for example for three terms with identical bases we obtain the following:

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

When we used the above law of exponents twice, we can also perform the same calculation for four terms in multiplication, five and so on..

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

$6b^3b^2b^{-2}=6b^{3+2-2}=6b^3$

Therefore, the correct answer is A.

Important note:

Here we need to emphasize that we should always ask the question - what does the exponent apply to?

For example, in this problem the exponent applies only to the $b$ bases and not to the numbers, more clearly, in the following expression: $5c^7$The exponent applies only to $c$ and not to the number 5,

whereas when we write: $(5c)^7$The exponent applies to each of the terms in the multiplication within the parentheses,

meaning:$(5c)^7=5^7c^7$

This is actually the application of the law of exponents:

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

Which follows both from the meaning of parentheses and from the definition of exponents.

Here we have multiplication between all the elements in the expression, so we will first use the commutative law in multiplication approach the numbers separately, for clarity we will approach it in stages:

$x^3\cdot7x\cdot2x^{-3}=7\cdot2\cdot x^3\cdot x\cdot x^{-3}=14\cdot x^3\cdot x\cdot x^{-3}$Note that it is possible (and even preferable) to skip the middle stage, meaning:

To write directly:$x^3\cdot7x\cdot2x^{-3}=14\cdot x^3\cdot x\cdot x^{-3}$

We will continue and use the associative law for multiplication between elements with the same bases:

$c^m\cdot c^n=c^{m+n}$Note that this law is also valid for several elements in multiplication and not just for two, for example for a multiplication of three elements with the same base we will get:

$c^m\cdot c^n\cdot c^k=c^{m+n}\cdot c^k=c^{m+n+k}$When can use the associative even for four, five, or more elements in a multiplication.

Let's go back to the problem, and apply the associative law:

$14x^3xx^{-3}=14x^{3+1-3}=14x^1=14x$And therefore the correct answer is c.

Important note:

Here it is necessary to emphasize that you always need to ask the question - what do the parentheses apply to?

For example, in the problem here the parentheses only apply to the bases of the-

$x$$$and not to the exponents, in a clearer way, also in the following expression:

$5c^7$The parentheses apply only to $c$and not to the exponent 5, as opposed to that when writing:

$(5c)^7$The parentheses apply to each of the multiplication elements within the parentheses, meaning:

$(5c)^7=5^7c^7$This is actually the application of the associative law:

$(w\cdot r)^n=w^n\cdot r^n$ resulting both from the meaning of the parentheses and from the definition of parentheses.

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, 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: $51x^{-3}y^{-1}$.

Given that there is a multiplication between all terms in the expression, we will apply the distributive property of multiplication. This allows us to handle the coefficients of terms raised to powers, as well as the terms themselves separately. For added clarity, let's handle it in steps:

$2a^2\cdot3a^4=2\cdot3\cdot a^2\cdot a^4=6\cdot a^2\cdot a^4$

Due to the multiplication between all terms we could do this, it should be noted that we can (and it's preferable to) skip the middle step, meaning:

Write directly:$2a^2\cdot3a^4=6\cdot a^2\cdot a^4$

From here on we will no longer write the multiplication sign. We will instead place the terms next to each other\ place the term next to its coefficient to indicate multiplication between them,

Proceed to apply the law of exponents for multiplication of terms with identical bases:

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

Note that this law applies to any number of terms being multiplied and not just two, for example for three terms with identical bases we obtain:

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

Whilst we used the law of exponents for two terms we can equally perform the same calculation for four terms or five and so on..,

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

$6a^2a^4=6a^{2+4}=6a^6$

Therefore the correct answer is C.

Important note:

Here we need to emphasize that we should always ask the question - what is the exponent being applied to?

For example, in this problem the exponent applies only to the bases of

$a$$$and not to the numbers, more clearly, in the following expression: $5c^7$the exponent applies only to $c$ and not to the number 5,

whereas when we write:$(5c)^7$the exponent applies to each term of the multiplication inside the parentheses,

meaning:$(5c)^7=5^7c^7$This is actually the application of the law of exponents:

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

which follows both from the meaning of parentheses and from the definition of exponents.

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.

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 $\mathbf{8xy^5}$.