Applying Combined Exponents Rules: Using multiple rules

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.

First, we'll enter the same fraction using the multiplication law between fractions, by 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, but use the conventional writing method where placing terms next to each other means multiplication.

Now we'll notice that both in the numerator and denominator, multiplication is performed between terms with identical bases, therefore we'll use 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 multiplication in the numerator and denominator:

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

where in the last step we calculated the sum of the exponents.

Now, we'll notice that we need to perform division (fraction=division operation between numerator and denominator) between terms with identical bases, therefore we'll use 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:

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

where in the last step we calculated the result of subtraction in the exponent.

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

therefore the correct answer is D.

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

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

Next, we'll notice that both in the numerator and denominator, multiplication occurs between terms with identical bases, so we'll use 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 here on, we will no longer indicate the multiplication sign, but use the conventional writing method where placing terms next to each other means multiplication.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

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

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

Now we notice that division is required between two terms with identical bases, so we'll use the power law for division between terms with identical bases:

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

We emphasize that 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:

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

where in the final step we calculated the subtraction between the exponents.

We have obtained the most simplified expression and therefore we are done.

Therefore, the correct answer is C.

$(\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}$

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:

$1$

Let's reduce the given expression step-by-step using the laws of exponents.

The expression to simplify is:

$\frac{\left(5^2\times2^3\times3\right)^3\times3^2}{2^4\times5^3}$

First, simplify the expression inside the bracket:

• Apply the power of a power rule $(a^m)^n = a^{m \times n}$ to each term:

  • $(5^2)^3 = 5^{2 \times 3} = 5^6$

  • $(2^3)^3 = 2^{3 \times 3} = 2^9$

  • $(3^1)^3 = 3^{1 \times 3} = 3^3$

• Substitute back into the expression:

• $\frac{5^6 \times 2^9 \times 3^3 \times 3^2}{2^4 \times 5^3}$

Next, combine the powers in the numerator:

• Use the product of powers rule $a^m \times a^n = a^{m+n}$:

  • Combine the 3s: $3^3 \times 3^2 = 3^{3+2} = 3^5$

• The refined numerator is $5^6 \times 2^9 \times 3^5$.

Now, simplify the fraction using division of powers:

• For 5's: $\frac{5^6}{5^3} = 5^{6-3} = 5^3$

• For 2's: $\frac{2^9}{2^4} = 2^{9-4} = 2^5$

• 3's remain $3^5$, as they only appear in the numerator.

Therefore, the final expression is:

$\boxed{5^3 \times 2^5 \times 3^5}$

To solve the problem of reducing the expression $\frac{\left(x^3\right)^2 \times y^5}{y^3}$, we'll follow these steps:

• Step 1: Simplify $\left(x^3\right)^2$ using the power of a power rule.

• Step 2: Simplify the expression using the division rule for $y^5$ divided by $y^3$.

Let's execute these steps:

Step 1: Apply the power of a power rule to $\left(x^3\right)^2$.

According to the power of a power rule, $(x^m)^n = x^{m \cdot n}$.

So, $\left(x^3\right)^2 = x^{3 \cdot 2} = x^6$.

Step 2: Simplify the division of exponents for the $y$ variable.

The expression now looks like $\frac{x^6 \times y^5}{y^3}$.

Using the division rule for exponents, $y^m / y^n = y^{m-n}$, we get:

$y^5 / y^3 = y^{5-3} = y^2$.

Final Expression: Combining the results from Step 1 and Step 2, we obtain:

$x^6 \times y^2$.

Therefore, the solution to the problem is $x^6 \times y^2$.

First, let's write the problem in an organized way and use 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:

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

Let's deal with the numerator, first using the law of exponents for multiplying terms with identical bases:

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

and we get:

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

Now let's use the law of exponents for 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 we got above:

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

Now let's use the law of exponents:

$a^0=1$

We can notice 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 get that 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, we got:

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

Let's handle each expression in the problem separately:

a. We'll start with the leftmost expression, first calculating the result of the multiplication in parentheses, and then use the power rule for power to a power:

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

Let's apply this to the problem for the first expression from the left:

$((7\cdot3)^2)^6=(21^2)^6=21^{2\cdot6}=21^{12}$

where in the final step we calculated the result of multiplication in the power expression,

We're done with this expression, let's move on to the next expression from the left.

b. Let's continue with the second expression from the left, using the power rule for power to a power that we mentioned above and apply it separately to each factor in this expression:

$(3^{-1})^3\cdot(2^3)^4=3^{-1\cdot3}\cdot2^{3\cdot4}=3^{-3}\cdot2^{12}$

Note that the multiplication factors we got have different bases, so we cannot further simplify this expression,

Therefore, let's combine parts a and b above in the result of the original problem:

$((7\cdot3)^2)^6+(3^{-1})^3\cdot(2^3)^4=21^{12}+3^{-3}\cdot2^{12}$

Therefore, the correct answer is answer d.

Let's start by handling the term in the multiplication that is in parentheses:

$(-y)^3$

For this, we'll recall the law of exponents for an exponent of a term in parentheses:

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

Accordingly, we get that:

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

We'll use this understanding in the problem and apply it to the aforementioned term:

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

where in the first stage we used the above understanding carefully - while using parentheses, and this is in order to remember that we're dealing with multiplication (not subtraction) and then we rearranged the expression using the distributive property of multiplication while remembering that a negative coefficient means multiplying by negative one,

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

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

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

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

where in 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 and recall 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 we got in the last stage:

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

where in the first stage we applied the above law of exponents carefully, because the term in the denominator has a negative exponent and then we simplified the expression in the exponent,

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

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

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 $y$, and this is in contrast to the understanding from the beginning of the solution where the entire expression: $-y$ is under the power of 3 because it's inside parentheses that are raised to the power of 3, meaning:

$(-y)^3$.

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 we'll use the fact that raising any number to the power of 0 gives the result 1, mathematically:

$X^0=1$

We'll apply this to both the numerator and denominator of the fraction in the problem:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1\cdot6^7}{36^4\cdot1}=\frac{6^7}{36^4}$

Next we'll note that -36 is a power of the number 6:

$36=6^2$

And we'll use this fact in the denominator to get expressions with identical bases in both the numerator and denominator:

$\frac{6^7}{36^4}=\frac{6^7}{(6^2)^4}$

Now we'll recall the power rule for power of a power to simplify the expression in the denominator:

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

And we'll also recall the power rule for division between terms with identical bases:

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

We'll apply these two rules to the expression we got above:

$\frac{6^7}{(6^2)^4}=\frac{6^7}{6^{2\cdot4}}=\frac{6^7}{6^8}=6^{7-8}=6^{-1}$

Where in the first stage we applied the first rule we mentioned earlier - the power of a power rule and simplified the expression in the exponent of the denominator term, then in the next stage we applied the second power rule mentioned before - the division rule for terms with identical bases, and again simplified the expression in the resulting exponent,

Finally we'll use the power rule for negative exponents:

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

And we'll apply it to the expression we got:

$6^{-1}=\frac{1}{6}$

Let's summarize everything we did, we got that:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1}{6}$

Therefore the correct answer is A.

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.

We use the formula:

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

$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$

We simplify:

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

$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$

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

• Step 1: Understand the expression $10^{3x}$.
• Step 2: Apply the power of a power rule to rewrite it.
• Step 3: Identify the correct equivalent expression from the options.

Now, let's work through each step:
Step 1: The expression given is $10^{3x}$, which involves a base of 10 and a combination of numerical and variable exponents, specifically $3x$.
Step 2: To rewrite this expression, we use the power of a power rule for exponents, which states $(a^m)^n = a^{m \cdot n}$. In our case, we want to reverse this process: we express $10^{3x}$ as $(10^3)^x$. Here, by viewing $3x$ as the product of $3$ and $x$, we can apply the rule effectively.
Step 3: We now compare our converted expression $(10^3)^x$ with the provided answer choices. The correct rewritten form is:
- Choice 3: $\left(10^3\right)^x$
Therefore, the solution to the problem is $\left(10^3\right)^x$. This matches the correct answer provided, validating our analysis and application of the power of a power rule.

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:

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

which 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:

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

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

We're done 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 apply this rule to the second term from the left:

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

and we're done 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.).

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

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

Let's apply this rule to the problem:

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

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

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

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

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

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

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

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

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

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

When we applied the above law of exponents carefully, this is because the term in the denominator has a negative exponent so we used parentheses,

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

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

Now let's recall the law of exponents for power to a power but in the opposite direction:

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

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

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

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

Therefore the correct answer is answer B.

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

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

Let's apply this law to the problem:

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

When we apply the above law of exponents to the second term from the left,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

To notice that:

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

Let's summarize the solution so far, we got that:

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

Therefore the correct answer is answer C.

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.