Power of a Product: Using multiple rules

Let's handle each term in the initial expression separately:

a. We'll start with the leftmost term, meaning the exponent on the multiplication in parentheses.

We'll use the power rule for exponents on multiplication in parentheses:

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

This rule states states that when an exponent applies to a multiplication in parentheses, it applies to each term in the multiplication when opening the parentheses.

Let's apply this to our problem for the leftmost term:

$(g\cdot a\cdot x)^4=g^4\cdot a^4\cdot x^4=g^4a^4x^4$

In the final step we dropped the multiplication sign and switched to the conventional multiplication notation by placing the terms next to each other.

Now that we're finished with the leftmost term, let's move on to the next term.

b. Let's continue with the second term from the left, using the power rule for exponents:

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

Let's now apply this rule to the second term from the left:

$(4^a)^x=4^{ax}$

Now are are finished with this term as well.

Let's summarize the results from a and b for the two terms in the initial expression:

$(g\cdot a\cdot x)^4+(4^a)^x=g^4a^4x^4+4^{ax}$

Therefore, the correct answer is c.

Notes:

a. For clarity and better explanation, in the solution above we handled each term separately. However, to develop proficiency and mastery in applying exponent rules, it is recommended to solve the problem as one unit from start to finish, where the separate treatment mentioned above can be done in the margin (or on a separate draft) if unsure about handling a specific term.

b. From the stated power rule for parentheses mentioned in solution a, it might seem that it only applies to two terms in parentheses, but in fact, it is valid for any number of terms in a multiplication within parentheses, as demonstrated in this problem and others.

It would be a good exercise to prove that if this rule is valid for exponents on multiplication of two terms in parentheses (as stated above), then it is also valid for exponents on multiplication of multiple terms in parentheses (for example - three terms, etc.).

Let's solve this in two stages. In the first stage, we'll use the rule for power of a product in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

This rule states that, when raising a product in parentheses to a power, the power applies to each factor of the product when opening the parentheses.

Let's apply this rule to our problem:

$(x^2\cdot y^3\cdot z^4)^2=(x^2)^2\cdot(y^3)^2\cdot(z^4)^2$

In our case, when expanding the parentheses, we apply the power to each factor of the product separately. However, given that each of the factors are raised to a power, we do this carefully and use parentheses.

Next, we'll use the power rule for a power of a power:

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

Let's apply this law to the expression we have:

$(x^2)^2\cdot(y^3)^2\cdot(z^4)^2=x^{2\cdot2}\cdot y^{3\cdot2}\cdot z^{4\cdot2}=x^4\cdot y^6\cdot z^8$

In the second stage we perform the multiplication operation in the power exponents of the factors we obtained.

Therefore, the correct answer is answer D.

Note:

From the above formulation of the power law for parentheses, it might seem that it only refers to two factors in a product within parentheses, but in fact it is valid for a power of a product of multiple factors in parentheses, as demonstrated in this problem.

It would be a good exercise to prove that if the above law is valid for a power of a product of two factors in parentheses (as it is formulated above), then it is also valid for a power of a product of multiple factors in parentheses (for example - three factors, etc.).

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

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

But before we perform the multiplication of fractions, note:

Important note-

Notice that in both fractions in the multiplication there are expressions that are binomials squared.

In this case, we're referring to:$(5x+4y)^3$ and$(5x+4y)^2$.

We'll treat these expressions simply as terms squared. This means that we won't open the parentheses and we'll treat both expressions exactly as we treat terms:

$a^3$and$a^2$.

Now let's return to the problem and continue from where we left off:

$$Let's apply the above-mentioned rule for multiplying fractions in the problem and perform the multiplication between the fractions:

$\frac{(5x+4y)^3\cdot7x}{45y\cdot x^2}\cdot\frac{3xy}{(5x+4y)^2}=\frac{7\cdot3xxy(5x+4y)^3}{45\cdot x^2y(5x+4y)^2}=\frac{7x^2y(5x+4y)^3}{15x^2y(5x+4y)^2}$

In the first stage we performed the multiplication between the fractions using the above rule, and in the second stage we reduced the numerical part in the resulting fraction; then we 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:

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

We applied this in the final stage to the numerator and denominator of the resulting fraction.

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

$\frac{7x^2y(5x+4y)^3}{15x^2y(5x+4y)^2}=\frac{7}{15}\cdot\frac{x^2}{x^2}\cdot\frac{y}{y}\cdot\frac{(5x+4y)^3}{(5x+4y)^2}=\frac{7}{15}\cdot1\cdot1\cdot\frac{(5x+4y)^3}{(5x+4y)^2}=\frac{7}{15}\cdot\frac{(5x+4y)^3}{(5x+4y)^2}$

Additionally, in the second stage, we applied the fact that dividing any number by itself gives 1.

We can now 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 have:

$\frac{7}{15}\cdot\frac{(5x+4y)^3}{(5x+4y)^2}=\frac{7}{15}\cdot(5x+4y)^{3-2}=\frac{7}{15}\cdot(5x+4y)^{1}=\frac{7}{15}\cdot(5x+4y)$

In the first stage, we applied the above law of exponents, treating the binomial as noted above, and then simplified the resulting expression remembering that raising a number to the power of 1 gives the number itself.

Let's summarize the solution to the problem:

$\frac{(5x+4y)^3\cdot7x}{45y\cdot x^2}\cdot\frac{3xy}{(5x+4y)^2}= \frac{7}{15}\cdot\frac{(5x+4y)^3}{(5x+4y)^2} =\frac{7}{15}\cdot(5x+4y)$

Therefore the correct answer is answer D.

Another 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 reduction of the numerical part to get directly the last line that we got:

$\frac{(5x+4y)^3\cdot7x}{45y\cdot x^2}\cdot\frac{3xy}{(5x+4y)^2}=\frac{7}{15}\cdot(5x+4y)$

(meaning we could have skipped the part where we expressed the fraction as a multiplication of fractions and even the initial multiplication of fractions we performed and gone straight to reducing 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 of the fractions in the problem, and also between the fractions themselves, multiplication is performed, meaning, that we can combine into one unified fraction as we did at the beginning and can apply the distributive property of multiplication and express as multiplication of fractions as mentioned above, etc., this is a point worth noting, since not in every problem we encounter all the conditions mentioned here in this note are met.

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:

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

But before we perform the fraction multiplication, let's note:

Important Note-

Notice that in both fractions of the multiplication there are expressions that are binomials squared.

In this case, we're referring to-$(5+3x)^8$and $(5+3x)^6$.

We'll treat these expressions simply as terms squared. This means that we won't open the parentheses and we'll treat both expressions exactly as we treat terms:

$a^8$and$a^6$

Now let's return to the problem and continue from where we left off:

$$Let's apply the rule for fraction multiplication mentioned above in the problem and perform the multiplication between the fractions:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}=\frac{136xy^5(5+3x)^8}{3xy^2y(5+3x)^6}=\frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6}$

In the first stage we performed the multiplication between the fractions using the above rule, then we simplified the expression in the fraction's denominator 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 in the final stage to the fraction's denominator that we got.

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

$\frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6}=\frac{136}{3}\cdot\frac{x}{x}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}\cdot1\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}$

Additionally in the second stage we applied the fact that dividing any number by itself gives 1.

We can now 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 that we got:

$\frac{136}{3}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}y^{5-3}(5+3x)^{8-6}=\frac{136}{3}y^2(5+3x)^2=45\frac{1}{3}\cdot y^2(5+3x)^2$

In the first stage we applied the above law of exponents, treating the binomial as noted above, and then simplified the resulting expression, in the final stage we converted the improper fraction we got to a mixed number.

Let's summarize the solution to the problem:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}= \frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6} =45\frac{1}{3}\cdot y^2(5+3x)^2$

Therefore the correct answer is answer B.

Another 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 numerical reduction to get directly the last line that we got:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}=45\frac{1}{3}\cdot y^2(5+3x)^2$

(meaning we could have skipped the part where we presented the fraction as a multiplication of fractions and even the initial fraction multiplication we performed and gone straight to reducing 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 it into a single fraction line like we did at the beginning and we can apply the distributive property and present as fraction multiplication as above, etc., this is a point worth noting, since not every problem we encounter will meet all the conditions mentioned here in this note.

First, we'll carefully open the parentheses, using two laws of exponents:

The first law is the exponent law that applies to parentheses containing multiplication of terms:

$(c\cdot z)^n=c^n\cdot z^n$which essentially states that when there is an exponent acting on parentheses containing multiplication between terms, when opening the parentheses the exponent will apply separately to each of the multiplication terms inside the parentheses.

The second law we'll use is the power of a power law:

$(c^m)^n=c^{m\cdot n}$which essentially states that when applying an exponent to a term that is already raised to a power (in the above form - inside parentheses for good order, but generally - also without the parentheses), we can interpret this as multiplication between the exponents within the exponent notation.

Let's return to the problem and first deal with the two parenthetical terms in the overall sum separately-

1. The first from left to right is:

$\big((8by)^3\big)^y=(8^3b^3y^3)^y=(8^3)^y(b^3)^y(y^3)^y=8^{3y}b^{3y}y^{3y}$where we used the first law above twice, first for the inner parentheses and then for the remaining parentheses, but we did this carefully because the terms in the multiplication within the parentheses are raised to powers, so we did this using additional parentheses, then we applied the power to the power (effectively opening the parentheses) using the second law above.

2. The second from left to right is:

$(3^x)^a=3^{xa}$where we applied the power to the power using the second law above.

Going back to the problem, we got:

$\big((8by)^3\big)^y+(3^x)^a=8^{3y}b^{3y}y^{3y} +3^{xa}$where we used 1 and 2 that we noted above.

We got the most simplified expression, so we're done.

Therefore, the correct answer is A.

Important note:

It's worth understanding the reason for the power of a power law mentioned above (the second law), this law comes directly from the definition of exponents:

$(c^m)^n=c^m\cdot c^m\cdot\ldots\cdot c^m=c^{m+m+m+\cdots+m}=c^{m\cdot n}$where in the first stage we applied the definition of exponents to the term in parentheses and multiplied it by itself n times, then we applied the law of exponents for multiplication between terms with identical bases mentioned above and interpreted the multiplication between the terms as a sum in the exponent notation,

Then we used the simple multiplication definition that states that if we connect a number to itself n times we can simply write it as multiplication, meaning:

$m+m+\cdots+m=m\cdot n$and therefore we get that:

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

In order to solve the problem we must use two power laws, as shown below:

A. Power property for terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$B. Power property for an exponent raised to another exponent:

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

We will apply these two power laws to the problem in two steps:

Let's start by applying the power law specified in A to the second term from the left in the given problem:

$\frac{2^7}{2^4}=2^{7-4}=2^3$In the first step we apply the power law specified in A and then proceed to simplify the resulting expression,

We then advance to the next step and apply the power law specified in B to the third term from the left in the given problem :

$(8^2)^5=8^{2\cdot5}=8^{10}$In the first stage we apply the power law specified in B and then proceed to simplify the resulting expression,

Let's summarize the two steps listed above to solve the general problem:

$(4\cdot7)^9+\frac{2^7}{2^4}+(8^2)^5= (4\cdot7)^9+2^3+8^{10}$In the final step, we calculate the result of multiplying the terms within the parentheses in the first term from the left:

$(4\cdot7)^9+2^3+8^{10}=28^9+2^3+8^{10}$Therefore, the correct answer is option c.

In solving the problem, we use two laws of exponents, which we will mention:

a. The law of exponents for multiplying powers with the same bases:

$a^m\cdot a^n=a^{m+n}$b. The law of exponents for a power of a power:

$(a^m)^n=a^{m\cdot n}$We will apply these two laws of exponents in solving the problem in two steps:

Let's start by applying the law of exponents mentioned in a' to the first expression on the left side of the problem:

$10^{-3}\cdot10^4=10^{-3+4}=10^1=10$When in the first step we applied the law of exponents mentioned in a' and in the following steps we simplified the expression that was obtained,

We continue to the next step and apply the law of exponents mentioned in b' and handle the third expression on the left side of the problem:

$(4^2)^5=4^{2\cdot5}=4^{10}$When in the first step we applied the law of exponents mentioned in b' and in the following steps we simplified the expression that was obtained,

We combine the two steps detailed above to the complete problem solution:

$10^{-3}\cdot10^4-(7\cdot9\cdot5)^3+(4^2)^5= 10-(7\cdot9\cdot5)^3+4^{10}$In the next step we calculate the result of multiplying the numbers inside the parentheses in the second expression on the left:

$10-(7\cdot9\cdot5)^3+4^{10}= 10-315^3+4^{10}$Therefore, the correct answer is answer b'.

Let's start by dealing with the multiplication term inside parentheses:

$(-y)^3$

For this, we must recall the rule for an exponent of a term inside parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

This gives us:

$(-y)^3=(-1\cdot y)^3=(-1)^3\cdot y^3=-1\cdot y^3=-y^3$

We'll now apply this to the aforementioned term:

$\frac{y^3\cdot y^{-4}\cdot(-y)^3}{y^{-3}}=\frac{y^3\cdot y^{-4}\cdot(-y^3)}{y^{-3}}=\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}}$

We rearranged the expression using the distributive property of multiplication while remembering that a negative coefficient means multiplying by negative one.

Next, we need to recall the law of exponents for multiplying terms with identical bases:

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

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

$\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}}=\frac{-y^{3+(-4)+3}}{y^{-3}}=\frac{-y^{3-4+3}}{y^{-3}}=-\frac{y^2}{y^{-3}}$

Iin the first stage we applied the above law of exponents to the multiplication terms (with identical bases) in the expression and in the final stage we remembered that negative one divided by negative one equals negative one.

Let's summarize the solution steps so far:

$\frac{y^3\cdot y^{-4}\cdot(-y)^3}{y^{-3}}=\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}} =-\frac{y^2}{y^{-3}}$

We'll continue by remembering the law of exponents for dividing terms with identical bases:

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

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

$-\frac{y^2}{y^{-3}}=-y^{2-(-3)}=-y^{2+3}=-y^5$

In the first stage we applied the above law of exponents carefully, because the term in the denominator has a negative exponent. Then we simplified the expression in the exponent.

Let's summarize the solution steps:

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

Therefore, the correct answer is answer A.

Note:

Let's note and emphasize that the minus sign in the final answer is not under the exponent, meaning the exponent doesn't apply to it but only to $y$. This is in contrast to the beginning of the solution where, for the entire expression, $-y$ is under the power of 3 because it's inside parentheses that are raised to the power of 3. Therefore:

$(-y)^3$.

In order to simplify the given expression, we will use the following two laws of exponents:

a. Law of exponents for multiplication of terms with identical bases:

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

b. Law of exponents for division of terms with identical bases:

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

Let's solve the given expression:

$\frac{2^3\cdot2^4}{2^5}=$First, since in the numerator we have multiplication of terms with identical bases, we'll use the law of exponents mentioned in a:

$\frac{2^3\cdot2^4}{2^5}= \\ \frac{2^{3+4}}{2^5}=\\ \frac{2^{7}}{2^5}=\\$We'll continue, since we have division of terms with identical bases, we'll use the law of exponents mentioned in b:

$\frac{2^{7}}{2^5}=\\ 2^{7-5}=\\ 2^2=\\ \boxed{4}$

Let's summarize the simplification of the given expression:

$\frac{2^3\cdot2^4}{2^5}= \\ \frac{2^{7}}{2^5}=\\ 2^2=\\ \boxed{4}$

Therefore, the correct answer is answer d.

We'll use the power rule for a power:

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

We'll apply this rule to the expression in the problem in two stages:

$((9xyz)^3)^4+(a^y)^x= (9xyz)^{3\cdot4}+(a^y)^x=(9xyz)^{12}+a^{yx}$

In the first stage, we apply the above rule initially to the first term in the expression and then proceed to deal with the outer parentheses. We then simplify the expression in the exponent whilst simultaneously applying the power rule to the second term in the sum in the problem's expression.

We'll continue by recalling the rule for powers that applies to parentheses containing the multiplication of terms:

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

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

$(9xyz)^{12}+(a^y)^x =9^{12} x^{12} y^{12}z^{12}+a^{yx}$

We apply the aforementioned power rule to the first term in the sum in the expression that we obtained in the last stage, and apply the power on the parentheses to each of the multiplication terms inside the parentheses.

Let's summarize the solution steps so far:

$((9xyz)^3)^4+(a^y)^x=(9xyz)^{12}+a^{yx} =9^{12} x^{12} y^{12} z^{12}+a^{yx}$

Therefore the correct answer is answer D.

First, we'll carefully expand the parentheses, using two laws of exponents:

The first law is the exponent law that applies to parentheses containing multiplication of terms:

$(c\cdot b)^n=c^n\cdot b^n$

This law essentially states that when there is an exponent acting on parentheses containing multiplication between terms, when opening the parentheses the exponent will apply separately to each of the multiplication terms inside the parentheses.

The second law we'll use is the power of a power law:

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

This law states that when applying an exponent to a term that is already raised to a power we can interpret this as multiplication between the exponents within the exponent notation.

Let's return to the problem and first deal with the two parenthetical terms in the overall sum separately:

1. The second from left to right is:

$(xyz)^{\frac{1}{4}}=x^{\frac{1}{4}}y^{\frac{1}{4}}z^{\frac{1}{4}}$

We open the parentheses using the first law mentioned above, so that when opening the parentheses we apply the exponent to each of the multiplication terms inside the parentheses.

2. The first from left to right is:

$\big((5a)^2\big)^3=(5^2a^2)^3=(5^2)^3(a^2)^3=5^{2\cdot3}a^{2\cdot3}=5^6a^6$

We used the first law above twice, first for the inner parentheses and then for the remaining parentheses, but we did this carefully since the terms in the multiplication within the parentheses are raised to powers and therefore we performed this using additional parentheses; and then we applied the power to the power (while effectively opening the parentheses) using the second law above.

Going back to the problem, we have:

$\big((5a)^2\big)^3+(xyz)^{\frac{1}{4}}=5^6a^6+ x^{\frac{1}{4}}y^{\frac{1}{4}}z^{\frac{1}{4}}$

We use 1 and 2 that we noted above.

We now have the most simplified expression.

Therefore the correct answer is C.

Important note:

It's worth noting the reasoning for the power of a power law mentioned above (the second law). This law comes directly from the definition of exponents:

$(c^m)^n=c^m\cdot c^m\cdot\ldots\cdot c^m=c^{m+m+m+\cdots+m}=c^{m\cdot n}$

In the first stage we applied the definition of exponents to the term in parentheses and multiplied it by itself n times, then we applied the law of exponents for multiplication between terms with identical bases mentioned above and interpreted the multiplication between the terms as a sum in the exponent.

Then we used the simple multiplication definition that says if we connect a number to itself n times we can simply write this as multiplication, meaning:

$m+m+\cdots+m=m\cdot n$

Therefore:

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

When solving the following problem, we will use two laws of exponents:

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

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

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

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

We will apply these two laws to the expression in the problem in two stages:

We'll start by applying the law of exponents mentioned in a to the second term from the left in the expression:

$9^{-3}\cdot9^4=9^{-3+4}=9^1=9$

Then, we'll apply the law of exponents mentioned in a in the first stage and simplified the resulting expression.

We'll continue to the next stage and apply the law of exponents mentioned in b and deal with the third term from the left in the expression. We'll do this in two steps:

$((7^2)^5)^6=(7^2)^{5\cdot6}=7^{2\cdot5\cdot6}=7^{60}$

In the first step we apply the law of exponents mentioned in b and eliminate the outer parentheses. In the next step, we apply the same law of exponents again and eliminate the remaining parentheses. In the following steps, we simplify the resulting expression.

Let's summarize the two stages detailed above for the complete solution of the problem:

$(9\cdot7\cdot6)^3+9^{-3}\cdot9^4+((7^2)^5)^6+2^4 = (9\cdot7\cdot6)^3+9+7^{60}+2^4$

In the next step we'll calculate the result of multiplying the terms inside the parentheses in the leftmost term:

$(9\cdot7\cdot6)^3+9+7^{60}+2^4 =378^3+9+7^{60}+2^4$

Therefore the correct answer is answer d.