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.).

Let's solve the expression given as: $\left(2\times9\times6\right)^{-7}$.

We 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 have:
$\left(2\times9\times6\right)^{-7} = \frac{1}{\left(2\times9\times6\right)^7}$.

This confirms the given correct answer:
$\frac{1}{\left(2\times9\times6\right)^7}$

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.

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:

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

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.

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.).