Power of a Product: Calculating powers with negative exponents

We begin by using the power rule for parentheses.

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

That is, the power applied to a product inside parentheses is applied to each of the terms within when the parentheses are opened,

We apply the above rule to the given problem:

$(3\cdot2\cdot4\cdot6)^{-4}=3^{-4}\cdot2^{-4}\cdot4^{-4}\cdot6^{-4}$

Therefore, the correct answer is option d.

Note:

According to the formula of the power property inside parentheses mentioned above, it might seem as though it refers to only two terms of the product inside of the parentheses, but in reality, it is also valid for the power over a multiplication of many terms inside parentheses, as was seen above.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms inside parentheses (as formulated above), then it is also valid for a power over several terms of the product inside parentheses (for example - three terms, etc.).

To solve the expression $\left(5\times8\right)^{-5}$, we need to apply the rules of exponents related to negative exponents.

First, let's recall the rule for negative exponents: for any non-zero number $a$: $a^{-n} = \frac{1}{a^n}$.

Applying this rule to our expression$\left(5\times8\right)^{-5}$:

• The base $5\times8$ is a product of two numbers 5 and 8.

• The exponent is -5, which means we have a negative exponent.

• According to the property of negative exponents, we invert the base and change the sign of the exponent:

Thus, $\left(5\times8\right)^{-5} = \frac{1}{\left(5\times8\right)^5}$.

After applying the rule, we arrive at the expression $\frac{1}{(5 \times 8)^5}$, which matches the given solution.

We first need to apply the exponent rule for powers with negative exponents, specifically the rule for the power of a product which states that:
$\left(a \times b \times c \right)^{-n} = \frac{1}{(a \times b \times c)^n}$.

In this problem, we have three numbers multiplied inside the parentheses: 2, 9, and 6. The exponent is -7.

By applying the power of a product rule with a negative exponent here, we get:
$\left(2\times9\times6\right)^{-7} = \frac{1}{\left(2\times9\times6\right)^7}$.

Therefore the correct answer is:
$\frac{1}{\left(2\times9\times6\right)^7}$

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To solve the expression $\left(9\times7\times8\right)^{-8}$, we need to apply the power of a product rule combined with the rule for negative exponents. The rule states that $a^{-n} = \frac{1}{a^n}$. So, a negative exponent indicates a reciprocal.

According to the power of a product rule, if you have a product raised to a power, it is the same as each factor being raised to that power: $(a \times b)^n = a^n \times b^n$.

So, applying the negative exponent rule to the original expression:

• Given: $\left(9\times7\times8\right)^{-8}$.

• Convert the negative exponent to positive by taking the reciprocal: $\frac{1}{\left(9\times7\times8\right)^8}$.

The correct expression after applying these rules is:

$\frac{1}{(9\times7\times8)^8}$.

We have the expression $\left(10 \times 7 \times 9\right)^{-5}$.

According to the rule for negative exponents, which states that $a^{-n} = \frac{1}{a^n}$, this expression can be rewritten as the reciprocal:

$\left(10 \times 7 \times 9\right)^{-5} = \frac{1}{\left(10 \times 7 \times 9\right)^5}$.

Thus, the simplified expression is:

$\frac{1}{\left(10 \times 7 \times 9\right)^5}$.

Step 1: We begin by applying the power of a product rule to the expression $\left(4 \times 7\right)^{-2}$. According to this rule, $(ab)^n = a^n \times b^n$. Therefore, we have:

$\left(4 \times 7\right)^{-2} = 4^{-2} \times 7^{-2}$

Step 2: Next, we use the negative exponent rule, which states that $a^{-n} = \frac{1}{a^n}$. Applying this rule to both parts, we get:

$4^{-2} = \frac{1}{4^2}$ and $7^{-2} = \frac{1}{7^2}$

So, $4^{-2} \times 7^{-2} = \frac{1}{4^2} \times \frac{1}{7^2}$

By multiplying these fractions, we obtain:

$\frac{1}{4^2 \times 7^2}$

Therefore, the solution to the problem is $\frac{1}{4^2 \times 7^2}$.

Keep in mind - we could have used the rules in the other way around, first the negative exponent rule, and only then the product rule and the result would still be the same!

To solve the problem of simplifying $(6 \times 5)^{-3}$, we will proceed as follows:

• Step 1: Apply the power of a product rule, which states $(a \times b)^n = a^n \times b^n$. In our case, apply this to get $(6 \times 5)^{-3} = 6^{-3} \times 5^{-3}$.
• Step 2: Use the negative exponent rule, which is $a^{-n} = \frac{1}{a^n}$. Applying this to each term, we find:
• $6^{-3} = \frac{1}{6^3}$
• $5^{-3} = \frac{1}{5^3}$
• Step 3: Multiply the results from Step 2:
• $6^{-3} \times 5^{-3} = \left(\frac{1}{6^3}\right) \times \left(\frac{1}{5^3}\right) = \frac{1}{6^3 \times 5^3}$.

Therefore, the expression $(6 \times 5)^{-3}$ simplifies to $\frac{1}{6^3 \times 5^3}$.

The correct answer choice is:

$\frac{1}{6^3\times5^3}$

To solve this problem, we'll simplify $\left(2\times7\times3\right)^{-6}$ using exponent rules:

• Step 1: Apply the power of a product rule
  The expression is $(2 \times 7 \times 3)^{-6}$. By the power-of-a-product rule, we have: $(2 \times 7 \times 3)^{-6} = 2^{-6} \times 7^{-6} \times 3^{-6}$.
• Step 2: Convert using negative exponent rule
  Use $a^{-n} = \frac{1}{a^n}$, so: $2^{-6} \times 7^{-6} \times 3^{-6} = \frac{1}{2^6 \times 7^6 \times 3^6}$.

Comparing with the given choices:

• Choice 1 suggests $\frac{1}{2^{-6} \times 7^{-6} \times 3^{-6}}$ which is incorrect.
• Choice 2 is $\frac{1}{2^6 \times 7^6 \times 3^6}$, which matches our simplified expression.
• Choice 3 is $\frac{1}{\left(2\times7\times3\right)^6}$ which also matches because: $\frac{1}{(2 \times 7 \times 3)^6} = \frac{1}{2^6 \times 7^6 \times 3^6}$

Since both choices B and C match the simplified expression, we conclude that choice D (B+C are correct) is the correct answer.

Therefore, the correct answer to the problem, as per our solution, is "B+C are correct".

Let's solve the problem step-by-step to find the value of $(5 \times 6 \times 8)^{-9}$.

Step 1: Recognize that the expression $(5 \times 6 \times 8)^{-9}$ has a negative exponent. According to the negative exponent rule $a^{-b} = \frac{1}{a^b}$, we can write:

$(5 \times 6 \times 8)^{-9} = \frac{1}{(5 \times 6 \times 8)^9}$

Step 2: Next, apply the power of a product rule. This rule states that $(xyz)^n = x^n \times y^n \times z^n$. Therefore, apply this to $(5 \times 6 \times 8)^9$:

$(5 \times 6 \times 8)^9 = 5^9 \times 6^9 \times 8^9$

Step 3: Substitute back into the fraction obtained in Step 1:

$\frac{1}{(5 \times 6 \times 8)^9} = \frac{1}{5^9 \times 6^9 \times 8^9}$

This result is the fully simplified expression sought for the original problem.

Therefore, the expression $(5 \times 6 \times 8)^{-9}$ simplifies to $\frac{1}{5^9 \times 6^9 \times 8^9}$.

Start with $\left(12 \times 7 \times 9\right)^{-5}$.

Apply the power of a product property, which states $(xyz)^n = x^n \times y^n \times z^n$:
$\left(12 \times 7 \times 9\right)^{-5} = 12^{-5} \times 7^{-5} \times 9^{-5}$

Use the negative exponent property, $a^{-n} = \frac{1}{a^n}$:
$12^{-5} \times 7^{-5} \times 9^{-5} = \frac{1}{12^{5}} \times \frac{1}{7^{5}} \times \frac{1}{9^{5}}$

This results in:
$\frac{1}{12^5 \times 7^5 \times 9^5}$

Thus, the solution to the expression is $\frac{1}{12^5 \times 7^5 \times 9^5}$.

Keep in mind - we could have used the rules in the other way around, first the negative exponent rule, and only then the product rule and the result would still be the same!

To solve the problem, we must correctly apply the rules for exponents to distribute the given negative exponent across the factors in the expression:

The expression given is $\left(4 \times 5\right)^{-2}$. First, we apply the power of a product rule:

$\left(4 \times 5\right)^{-2} = 4^{-2} \times 5^{-2}$

This rule allows us to distribute the negative exponent $-2$ to both factors in the product inside the parentheses. Each factor is affected by the exponent.

Now, let’s verify this against the choices provided:

• Choice 3: $4^{-2} \times 5^{-2}$ matches our solution with proper application of the power of a product rule.

Therefore, the correct and corresponding expression is $\boxed{4^{-2}\times5^{-2}}$.

To solve this problem, we'll apply the power of a product rule for exponents:

• Step 1: Identify the expression $(9 \times 2)^{-6}$.
• Step 2: Apply the power of a product rule, which states $(a \times b)^n = a^n \times b^n$.
• Step 3: Simplify the expression using the rule $(9 \times 2)^{-6} = 9^{-6} \times 2^{-6}$.

By performing these steps, the expression $(9 \times 2)^{-6}$ becomes equivalent to $9^{-6} \times 2^{-6}$.

Therefore, the solution to the problem is $9^{-6} \times 2^{-6}$.

To solve the expression $(2 \times 7 \times 3)^{-3}$, let's proceed with the following steps:

Step 1: Recognize that we have a product inside the parentheses, $2 \times 7 \times 3$, which is raised to the power of $-3$.

Step 2: Apply the power of a product rule, which states that $(a \times b \times c)^n = a^n \times b^n \times c^n$. This gives us:

$(2 \times 7 \times 3)^{-3} = 2^{-3} \times 7^{-3} \times 3^{-3}$

Therefore, applying the exponent rule, we have:

$2^{-3} \times 7^{-3} \times 3^{-3}$

To solve this problem, we will apply the rule for the power of a product and handle the negative exponent:

• The expression given is $(9 \times 7 \times 8)^{-8}$.
• According to the Power of a Product rule: $(a \times b \times c)^n = a^n \times b^n \times c^n$ where $a$, $b$, and $c$ are the factors inside the parentheses, and $n$ is the exponent.
• Applying this rule, we distribute the exponent $-8$ to each of the factors inside the parentheses:
• $(9 \times 7 \times 8)^{-8} = 9^{-8} \times 7^{-8} \times 8^{-8}$.

Therefore, the solution to the problem is $9^{-8} \times 7^{-8} \times 8^{-8}$.

To solve the expression $(3 \times 7)^{-4}$, we need to apply the rules for negative exponents:

The expression $(3 \times 7)^{-4}$ can be rewritten using the negative exponent rule, which states that $x^{-n} = \frac{1}{x^n}$. Applying this rule gives:
$(3 \times 7)^{-4} = \frac{1}{(3 \times 7)^4}$

This simplifies the problem, as now it is expressed in terms of a positive exponent.

Checking our choices, the correct expression is match with choice 3: $\frac{1}{(3 \times 7)^4}$

Thus, $(3 \times 7)^{-4} = \frac{1}{(3 \times 7)^4}$.

We begin by applying the power rule to the products within the parentheses:

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

That is, the power applied to a product within parentheses is applied to each of the terms when the parentheses are opened,

We apply the rule to the given problem:

$(8\cdot9\cdot5\cdot3)^{-2}=8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2}$

Therefore, the correct answer is option c.

Note:

Whilst it could be understood that the above power rule applies only to two terms of the product within parentheses, in reality, it is also valid for the power over a multiplication of multiple terms within parentheses, as was seen in the above problem.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms within parentheses (as formulated above), then it is also valid for a power over several terms of the product within parentheses (for example - three terms, etc.).

Use the power property for a power in parentheses where there is a multiplication of its terms:

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

We apply this law to the problem expression:

$(5\cdot12\cdot4\cdot6)^{a+3bx}=5^{a+3bx}12^{a+3bx}4^{a+3bx}6^{a+3bx}$

When we apply a power to parentheses where its terms are multiplied, we do it separately and keep the multiplication.

Therefore, the correct answer is option d.

To solve this expression, we apply the rule for negative exponents.

The expression given is $\frac{1}{(2 \times 7 \times 8)^9}$. We recognize this as the reciprocal of a power, which can be rewritten using the negative exponent rule:

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

Thus, $\frac{1}{(2 \times 7 \times 8)^9}$ can be rewritten as $(2 \times 7 \times 8)^{-9}$.

Thus, the correct answer is choice 1: $(2 \times 7 \times 8)^{-9}$.

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

• Step 1: Recognize the given expression: $\frac{1}{(10 \times 7)^5}$.

• Step 2: Apply the rule of negative exponents, which states: $\frac{1}{a} = a^{-1}$.

• Step 3: Express the reciprocal with a negative exponent: $\frac{1}{(10 \times 7)^5} = (10 \times 7)^{-5}$.

Now, let's further simplify the expression:
Given that $(a \times b)^n = a^n \times b^n$, we can rewrite $(10 \times 7)^5$ as $10^5 \times 7^5$. Thus, $(10 \times 7)^{-5} = (10^5 \times 7^5)^{-1} = 10^{-5} \times 7^{-5} = (10 \times 7)^{-5}$.

Therefore, the solution to the problem in the expression form is: $\left(10 \times 7\right)^{-5}$.

To solve the problem, follow these steps:

• Given the expression $\frac{1}{8^{-7} \times 9^{-7} \times 5^{-7}}$.
• Apply the negative exponent rule: Each term in the denominator is raised to a negative power.
• Write each term with positive exponents using the reciprocal rule: $\frac{1}{a^{-n}} = a^n$.
• This results in the expression: $8^7 \times 9^7 \times 5^7$.
• The power of a product rule tells us that this is equivalent to $(8 \times 9 \times 5)^7$.

Therefore, the solution is $\left(8 \times 9 \times 5\right)^7$.

Comparing with the multiple-choice options provided, the correct choice is: $\left(8 \times 9 \times 5\right)^7$.