Powers of a Fraction: Number of terms

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

• Step 1: Identify the expression $\left(\frac{3 \times 5 \times 7}{4 \times 8 \times 10}\right)^5$.

• Step 2: Apply the rule for powers of fractions to simplify the expression.

• Step 3: Simplify the products and raise each factor to the power of 5.

Now, let's work through each step:
Step 1: The expression given is $\left(\frac{3 \times 5 \times 7}{4 \times 8 \times 10}\right)^5$.
Step 2: According to the power of a fraction rule, $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$, we can distribute the power of 5 to both the numerator and denominator:

$\left(\frac{3 \times 5 \times 7}{4 \times 8 \times 10}\right)^5 = \frac{(3 \times 5 \times 7)^5}{(4 \times 8 \times 10)^5}$

Step 3: Now apply the power to each factor:

$= \frac{3^5 \times 5^5 \times 7^5}{4^5 \times 8^5 \times 10^5}$

This matches the expression in choice 2. Therefore, the correct corresponding expression is $\frac{3^5 \times 5^5 \times 7^5}{4^5 \times 8^5 \times 10^5}$ and also $\frac{(3 \times 5 \times 7)^5}{(4 \times 8 \times 10)^5}$.

Therefore, the solution to the problem is A+B.

To solve the problem, follow these steps:

• Step 1: Use the property $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ to distribute the exponent 8 to both the numerator and the denominator.
  
• Step 2: Raise each component in the numerator and the denominator to the power of 8.

Start by applying the rule to the entire fraction:

$\left(\frac{11\times5\times4}{9\times13\times17}\right)^8 = \frac{(11\times5\times4)^8}{(9\times13\times17)^8}$

Then apply the rule of powers to the product inside both the numerator and denominator:

$= \frac{11^8 \times 5^8 \times 4^8}{9^8 \times 13^8 \times 17^8}$

Therefore, the solution to the given problem is:

$\frac{11^8\times5^8\times4^8}{9^8\times13^8\times17^8}$

To solve this expression, we need to apply the rules of exponents to simplify $\left(\frac{1}{4 \times 6 \times 9}\right)^{-4}$.

First, using the power of a fraction rule $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$, we express each term separately as follows:

$(4 \times 6 \times 9)^{-4} = 4^{-4} \times 6^{-4} \times 9^{-4}$

Therefore, the original negative exponent transforms to a multiplication of three positive exponents in the denominator represented as:

$\frac{1^{-4}}{4^{-4} \times 6^{-4} \times 9^{-4}}$

Thus, the corresponding expression in terms of powers with negative exponents is:

$\frac{1^{-4}}{4^{-4} \times 6^{-4} \times 9^{-4}}$

The correct answer is Choice 4.

To solve this problem, let's break down the expression $\left(\frac{3}{5 \times 8 \times 7}\right)^{-2}$:

• Step 1: Recognize that we have a fraction raised to a negative exponent.
• Step 2: Apply the power of a fraction rule: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$.
• Step 3: Specifically, apply the rule with a negative exponent: $\left(\frac{3}{5 \times 8 \times 7}\right)^{-2} = \frac{3^{-2}}{(5 \times 8 \times 7)^{-2}}$.
• Step 4: Use the negative exponent rule for each element in the expression $(5 \times 8 \times 7)^{-2} = 5^{-2} \times 8^{-2} \times 7^{-2}$.

This gives us the expression: $\frac{3^{-2}}{5^{-2} \times 8^{-2} \times 7^{-2}}$.

Therefore, the correct expression is $\frac{3^{-2}}{5^{-2} \times 8^{-2} \times 7^{-2}}$.

The given expression is:
$\left(\frac{7\times11\times19}{3\times12\times15}\right)^{-4}$

To solve this expression, we need to apply the rules of exponents, specifically the rule for powers of a fraction. For any fraction$\left(\frac{a}{b}\right)^{-n}$, the expression is equivalent to$\left(\frac{b}{a}\right)^n$.
Therefore, negative exponents indicate that the fraction should be flipped and raised to the positive of that exponent.

Substitute the terms into this formula:
1. Flip the fraction: $\left(\frac{3\times12\times15}{7\times11\times19}\right)$
2. Raise both numerator and denominator to the power of 4:
Thus, we have:
$\left(\frac{3\times12\times15}{7\times11\times19}\right)^{4}$

Now evaluating each term individually:
- In the numerator:
- $3^4\times12^4\times15^4$
- In the denominator:
- $7^4\times11^4\times19^4$

Applying the negative exponent rule, each individual factor in both numerator and denominator should be inverted, altering the exponents to negative:
1. Numerator becomes: $3^{-4}\times12^{-4}\times15^{-4}$
2. Denominator becomes: $7^{-4}\times11^{-4}\times19^{-4}$

Rewriting the expression, we achieve:
$\frac{7^{-4}\times11^{-4}\times19^{-4}}{3^{-4}\times12^{-4}\times15^{-4}}$

This matches precisely the provided solution.

The solution to the question is:$\frac{7^{-4}\times11^{-4}\times19^{-4}}{3^{-4}\times12^{-4}\times15^{-4}}$

To solve the problem, let's follow these steps:

• Step 1: Recognize that the given expression is $\left(\frac{10 \times 3}{7 \times 9}\right)^{-4}$. A negative exponent indicates that we should take the reciprocal of the base.
• Step 2: Rewrite this expression using the negative exponent rule: $\left(\frac{10 \times 3}{7 \times 9}\right)^{-4} = \left(\frac{7 \times 9}{10 \times 3}\right)^{4}$ This step inverts the fraction and changes the exponent from $-4$ to $4$.
• Step 3: Apply the exponent to each component of the fraction: $\left(\frac{7 \times 9}{10 \times 3}\right)^{4} = \frac{(7 \times 9)^{4}}{(10 \times 3)^{4}}$ This separates the powers between the numerator and the denominator.
• Step 4: Distribute the powers inside each product: $= \frac{7^4 \times 9^4}{10^4 \times 3^4}$ This is achieved by applying $(ab)^n = a^n \times b^n$ to both the numerator and the denominator.

Therefore, the simplified expression is $\frac{7^4 \times 9^4}{10^4 \times 3^4}$, which corresponds to choice 3 in the provided answer choices.

To solve the problem, we need to find the equivalent expression for the given negative exponent:

Step 1: Identify the base fraction as $\frac{6\times8}{2\times7}$.

Step 2: Apply the negative exponent rule:

The expression $\left(\frac{6\times8}{2\times7}\right)^{-5}$ can be rewritten using the property that $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$.

Thus, we have:

$\left(\frac{6\times8}{2\times7}\right)^{-5} = \left(\frac{2\times7}{6\times8}\right)^5$

Therefore, the correct equivalent expression is:

$\left(\frac{2\times7}{6\times8}\right)^5$

Hence, the choice corresponding to this expression is correct.

Therefore, the solution to the problem is $\left(\frac{2\times7}{6\times8}\right)^5$, which corresponds to choice 1.

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

• Step 1: Apply the negative exponent rule.
• Step 2: Use the power of a fraction rule to simplify the expression.
• Step 3: Ensure calculations are conducted correctly, and choose the matching answer from the choices.

Now, let's work through each step:
Step 1: The initial expression is $\left(\frac{3 \times 7}{4 \times 6}\right)^{-6}$. First, use the negative exponent rule: $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$. So, the expression becomes $\left(\frac{4 \times 6}{3 \times 7}\right)^{6}$.

Step 2: Apply the power of a fraction rule: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$. Thus, the expression can be re-written as:

$\frac{(4 \times 6)^6}{(3 \times 7)^6}$

Step 3: Assess the provided answer choices and determine which one matches our derived expression. Choice 3, $\frac{(4 \times 6)^6}{(3 \times 7)^6}$, correctly corresponds to our simplified result.

Therefore, the solution to the problem is $\frac{(4 \times 6)^6}{(3 \times 7)^6}$, which corresponds to choice 3.

Let's solve the problem step-by-step:

The given expression is $\left(\frac{4}{9 \times 7}\right)^{-3}$.

Step 1: Apply the negative exponent rule. A negative exponent $-n$ can be transformed by reciprocal and changing the sign of the exponent. Therefore, $\left(\frac{4}{9 \times 7}\right)^{-3} = \left(\frac{9 \times 7}{4}\right)^3$.

Step 2: Express the denominator as a product of integers for clarity: $\frac{9 \times 7}{4}$ is clearer than $\frac{63}{4}$ in context for further steps.

Step 3: Apply the power of a fraction rule. Where $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$, therefore, $\left(\frac{9 \times 7}{4}\right)^3 = \frac{(9 \times 7)^3}{4^3}$.

Step 4: Separate the powers within the fraction: The expression $(9 \times 7)^3$ can be expanded using individual exponents: $(9 \times 7)^3 = 9^3 \times 7^3$.

Thus, our expression simplifies to: $\frac{9^3 \times 7^3}{4^3}$.

After analyzing the problem and solving it using the rules of exponents, we conclude that the correct expression is $\frac{9^3 \times 7^3}{4^3}$.

Therefore, the correct choice from the provided options is choice 2: $\frac{9^3 \times 7^3}{4^3}$.

To solve this problem, let's begin by applying the mathematical rules for negative exponents and exponents of a fraction.

Step 1: Apply the negative exponent rule:

• Given: $\left(\frac{11 \times 9}{4}\right)^{-5}$.

• Using the rule $(a/b)^{-n} = (b/a)^n$, we rewrite this as $\left(\frac{4}{11 \times 9}\right)^5$.

Step 2: Simplify the expression:

• Analyzing the expression $\left(\frac{4}{11 \times 9}\right)^5$, we see that this is equivalent to:

• $\frac{4^5}{(11 \times 9)^5}$.

• Notice that $(11 \times 9)^5$ can also be written as $11^5 \times 9^5$ using properties of exponents.

• Thus, $\frac{4^5}{11^5 \times 9^5}$ is another way to express this fraction.

Step 3: Compare with given choices:

• Choice 2: $\frac{4^5}{11^5 \times 9^5}$ matches our final expression.

• Notice also Choice 3: $\frac{4^5}{(11 \times 9)^5}$ matches the form before simplifying the denominator completely to separate power terms.

Therefore, after comparison, Options B and C are indeed correct and thus the correct response is: B+C are correct.

To solve this problem, we will follow the standard rule for evaluating expressions with negative exponents:

• Step 1: Identify the original expression – $\left(\frac{4 \times 5}{3 \times 2}\right)^{-2}$.
• Step 2: Apply the negative exponent rule which states $(a/b)^{-n} = (b/a)^{n}$.
• Step 3: Simplify the expression.

Now, let's proceed through each step:

Step 1: The expression within the parentheses is $\frac{4\times5}{3\times2}$, which simplifies to $\frac{20}{6}$ or further reduced to $\frac{10}{3}$. However, for the purpose of matching the choices given, we'll use the product form directly.

Step 2: Apply the negative exponent formula:
$\left(\frac{4 \times 5}{3 \times 2}\right)^{-2} = \left(\frac{3 \times 2}{4 \times 5}\right)^2$

Step 3: The result of this step shows the reciprocal raised to the power of 2:

The solution to the problem is therefore represented as:
$\left(\frac{3 \times 2}{4 \times 5}\right)^2$.

Upon reviewing the choices provided, this corresponds to choice 1.

To simplify the expression $\left(\frac{3\times7}{5\times8}\right)^{-3}$, we follow these steps:

• Step 1: Apply the rule for negative exponents, which states that $a^{-b} = \frac{1}{a^b}$. For fractions, $\left(\frac{a}{b}\right)^{-n} = \frac{b^n}{a^n}$.
• Step 2: Rewrite the expression by applying this rule:
  $\left(\frac{3\times7}{5\times8}\right)^{-3} = \frac{(5\times8)^3}{(3\times7)^3}$
• Step 3: Simplify the expression by recognizing the bases to the power of 3: $\frac{(5\times8)^3}{(3\times7)^3}$

The expression can be left as is because it matches one of the choices provided. Performing further calculations here is unnecessary since we have correctly applied the negative exponent rule.

Therefore, the correct simplified form of the expression is $\frac{(5\times8)^3}{(3\times7)^3}$, which corresponds to choice 2.

First, let's note that the first term in the multiplication in the problem has an exponent of 0, and any number (different from zero) raised to the power of zero equals 1, meaning:

$X^0=1$Therefore, we get that the expression in the problem is:

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= 1\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}=(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}$

Later, we will use the law of exponents for negative exponents:

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

And before we proceed to solve the problem let's understand this law in a slightly different, indirect way:

Let's note that if we treat this law as an equation (which it indeed is in every way), and multiply both sides of the equation by the common denominator which is:

$a^n$we get:

$a^{-n}=\frac{1}{a^n}\\ \frac{a^n}{1} =\frac{1}{a^n}\hspace{8pt} \text{/}\cdot a^n\\ a^n\cdot a^{-n}=1$

Where in the first stage we remembered that any number can be represented as itself divided by 1, we applied this to the left side of the equation, then we multiplied by the common denominator and to know by how much we multiplied each numerator (after reduction with the common denominator) we addressed the question "by how much did we multiply the current denominator to get the common denominator?".

Let's look at the result we got:

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

Meaning that $a^n,\hspace{4pt}a^{-n}$are reciprocal numbers, or in other words:

$a^n$ is reciprocal to $a^{-n}$ (and vice versa),

and specifically:

$a,\hspace{4pt}a^{-1}$ are reciprocal to each other,

We can apply this understanding to the problem if we also remember that the reciprocal of a fraction is obtained by switching the numerator and denominator, meaning that the fractions:

$\frac{a}{b},\hspace{4pt}\frac{b}{a}$ are reciprocal fractions - which can be easily understood, since their multiplication will clearly give the result 1,

And if we combine this with the previous understanding, we can easily conclude that:

$\big(\frac{a}{b}\big)^{-1}=\frac{b}{a}$

In other words, raising a fraction to the power of negative one will give a result that is the reciprocal fraction, obtained by switching the numerator and denominator.

Let's return to the problem, the expression we got in the last stage is:

$(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}$

We'll use what was explained earlier and note that the fraction in parentheses in the second term of the multiplication is the reciprocal fraction to the fraction in parentheses in the first term of the multiplication, meaning that:$\frac{13}{2}= \big(\frac{2}{13} \big)^{-1}$Therefore we can simply calculate the expression we got in the last stage by converting to a common base using the above understanding:

$(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} = (\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5}$

Where we actually just replaced the fraction in parentheses in the second term of the multiplication with its reciprocal raised to the power of negative one as mentioned earlier,

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

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

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

$(\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{(-1)\cdot(-5)}=(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}$

Where in the first stage we applied the above law of exponents and then simplified the expression that resulted,

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

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= (\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} = (\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}$

In the next stage we'll recall the law of exponents for multiplication of terms with identical bases:

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

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

$(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5} =(\frac{2}{13}\big)^{-2+5}=(\frac{2}{13}\big)^{3}$

Where in the first stage we applied the above law of exponents and then simplified the expression,

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

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= (\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}=(\frac{2}{13}\big)^{3}$

Therefore the correct answer is answer A.

To solve this problem, we'll use the rule for negative exponents. Given the expression:

$\left(\frac{4\times8}{5\times17\times3}\right)^{-4}$

Step 1: Apply the property of exponents $(a/b)^{-n} = (b/a)^n$ to change the negative exponent to a positive one. Therefore, we take the reciprocal of the fraction:

$\left(\frac{5\times17\times3}{4\times8}\right)^4$

Therefore, the rewritten expression is $\left(\frac{5\times17\times3}{4\times8}\right)^4$.

Comparing this with the given answer choices, we find:

• Choice 1: $\left(\frac{5\times17\times3}{4\times8}\right)^4$ - Correct

Thus, the solution to the problem is $\left(\frac{5\times17\times3}{4\times8}\right)^4$.

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

• Step 1: Identify the original fraction inside the expression, which is $\frac{7 \times 13}{6 \times 10 \times 12}$.
• Step 2: Apply the negative exponent rule, which states $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$. This means we need to find the reciprocal of the fraction and change the sign of the exponent to positive.

Now, let's work through these steps:

Step 1: The problem gives us the expression $\left(\frac{7 \times 13}{6 \times 10 \times 12}\right)^{-3}$.

Step 2: Using the reciprocal rule for negative exponents, we will rewrite the given expression by taking the reciprocal of the fraction:
$\left(\frac{7 \times 13}{6 \times 10 \times 12}\right)^{-3} = \left(\frac{6 \times 10 \times 12}{7 \times 13}\right)^{3}$

By applying the rule, the expression becomes $\left(\frac{6 \times 10 \times 12}{7 \times 13}\right)^3$, which is the required positive exponent form.

Therefore, the correct expression corresponding to the problem is $\left(\frac{6 \times 10 \times 12}{7 \times 13}\right)^3$.

To solve this problem, we'll apply the mathematical rules for exponents and fractions:

• Step 1: Identify the structure: The given expression is $\left(\frac{3 \times 5 \times 7}{2 \times 4 \times 6}\right)^{-2}$.

• Step 2: Apply the exponent rule by flipping the fraction due to the negative exponent: $\left(\frac{3 \times 5 \times 7}{2 \times 4 \times 6}\right)^{-2} = \left(\frac{2 \times 4 \times 6}{3 \times 5 \times 7}\right)^{2}$

• Step 3: Rewrite using the power of a product rule: $\left(\frac{2 \times 4 \times 6}{3 \times 5 \times 7}\right)^{2} = \frac{(2 \times 4 \times 6)^2}{(3 \times 5 \times 7)^2}$

• Step 4: Recognize that this matches the form described by choices. Compare with options: - Option 2: $\frac{2^2 \times 4^2 \times 6^2}{3^2 \times 5^2 \times 7^2}$ - Option 3: $\frac{(2 \times 4 \times 6)^2}{(3 \times 5 \times 7)^2}$ is similar since each factor is squared, aligning with separate sea-lined approaches by recombining within parentheses.

Therefore, Option 2 and Option 3 both correctly represent the expression and the answer is: B+C are correct.

To solve this problem, we'll apply the rules for negative exponents:

We start with the expression: $\left(\frac{7 \times 11 \times 19}{3 \times 12 \times 15}\right)^{-6}$.

By the rule for negative exponents, $(a/b)^{-n} = (b/a)^n$, we invert the fraction and change the exponent from $-6$ to $6$:

$\left(\frac{3 \times 12 \times 15}{7 \times 11 \times 19}\right)^6$

This transformation capitalizes on the idea that negative exponents reflect a reciprocal relationship.

Therefore, the expression with positive exponents is: $\left(\frac{3 \times 12 \times 15}{7 \times 11 \times 19}\right)^6$.

We are given the expression: $\left(\frac{1}{2\times3\times4}\right)^{-2}$. We need to simplify it using the rules of exponents.

• Step 1: Identify the base of the exponent.
  The base is $\frac{1}{2\times3\times4}$.
  

• Step 2: Apply the rule for negative exponents.
  For a fraction $\frac{1}{a}$ with a negative exponent, $\left( \frac{1}{a} \right)^{-n} = a^n$. Therefore, $\left(\frac{1}{2\times3\times4}\right)^{-2} = (2\times3\times4)^2$.
  

• Step 3: Expand the expression.
  $(2\times3\times4)^2 = 2^2 \times 3^2 \times 4^2$.

Thus, the simplified expression is: $2^2\times3^2\times4^2$

To solve the expression $\left(\frac{1}{5\times6\times7}\right)^{-3}$, follow these steps:

• Step 1: Recognize that the expression $\left(\frac{1}{5\times6\times7}\right)^{-3}$ has a negative exponent.
• Step 2: Use the rule for negative exponents: $\left(\frac{1}{a}\right)^{-n} = a^{n}$.
• Step 3: Apply the rule: $\left(\frac{1}{5\times6\times7}\right)^{-3} = \left(5\times6\times7\right)^{3}$.

Therefore, the expression simplifies to $\left(5\times6\times7\right)^3$.

The correct answer is $\left(5\times6\times7\right)^3$.

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

• Step 1: Identify the negative exponent and use the rule $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$.
• Step 2: Apply the rule to the expression $\left(\frac{1}{4 \times 6 \times 9}\right)^{-4}$.
• Step 3: Simplify the result.

Now, let's work through each step:
Step 1: The given expression is $\left(\frac{1}{4 \times 6 \times 9}\right)^{-4}$. Notice the negative exponent $-4$.
Step 2: According to the rule, flipping the fraction and changing the sign of the exponent, we get $\left(4 \times 6 \times 9\right)^{4}$.
Step 3: Thus, the expression simplifies to $4^4 \times 6^4 \times 9^4$.

Therefore, the solution to the problem is $4^4 \times 6^4 \times 9^4$, which corresponds to choice 2.