Negative Exponents: Using the laws of exponents

$(8\times9\times5\times3)^{-2}=$

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.).

$8^{-2}\times9^{-2}\times5^{-2}\times3^{-2}$

$(\frac{2}{3})^{-4}=\text{?}$

We use the formula:

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

Therefore, we obtain:

$(\frac{3}{2})^4$

We use the formula:

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

Therefore, we obtain:

$\frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}$

$\frac{81}{16}$

$\frac{10^4\cdot0.1^{-3}\cdot10^{-8}}{1000}=\text{?}$

$0.0001$

$\frac{2^{-4}\cdot(\frac{1}{2})^8\cdot2^{10}}{2^3}=\text{?}$

$2^{-5}$

$9^4\cdot3^{-8}\cdot\frac{1}{3}=\text{?}$

$3^{-1}$

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

$5(5^3-1)$

$45^{-80}\cdot\frac{1}{45^{-81}}\cdot49\cdot7^{-5}=\text{?}$

$\frac{45}{7^3}$

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

$(3^4)^x$

$10^8+10^{-4}+(\frac{1}{10})^{-16}=\text{?}$

$10^8+\frac{1}{10^4}+10^{16}$

$\frac{7^8}{7^{-4}\cdot4^2}\cdot32=\text{?}$

$2\cdot7^{12}$

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{81}=\text{?}$

$(\frac{2}{3})^3$

$7^2\cdot(3^5)^{-1}\cdot\frac{1}{4}\cdot\frac{1}{3^2}=\text{?}$

$\frac{4^{-1}3^{-7}}{7^{-2}}$