Question

exponents and radicals
student activity sheet 1; overview

1. reinforce create expressions that meet the conditions specified.

a. the expression simplifies to $4^2 x^4 y$ using the multiplication rule.
b. the expression simplifies to $\frac{5x^7}{y^3}$ using the division rule, and has at least one negative exponent.
c. the expression simplifies to $5^3 x^5 y^2$ using the multiplication rule and the power rule, and has at least two negative exponents.

Explanation:

Part a

Step 1: Recall exponent multiplication rule

The multiplication rule for exponents is \(a^m \cdot a^n = a^{m + n}\). We need to create an expression that simplifies to \(4^{2}x^{7}y^{4}\) using this rule. Let's break down the exponents for \(x\) and \(y\) into sums. For \(x^{7}\), we can write \(7 = 3+4\), and for \(y^{4}\), we can write \(4 = 2 + 2\). Also, \(4^{2}\) can be part of the expression.
Let's construct the expression: \(4^{2}x^{3}y^{2}\cdot x^{4}y^{2}\)

Step 2: Apply the multiplication rule

Using \(a^m \cdot a^n=a^{m + n}\) for \(x\) terms: \(x^{3}\cdot x^{4}=x^{3 + 4}=x^{7}\)
For \(y\) terms: \(y^{2}\cdot y^{2}=y^{2+2}=y^{4}\)
And the coefficient term: \(4^{2}\) remains as it is. So the product \(4^{2}x^{3}y^{2}\cdot x^{4}y^{2}=4^{2}x^{7}y^{4}\), which matches the required simplified form.

Step 1: Recall exponent division rule

The division rule for exponents is \(\frac{a^m}{a^n}=a^{m - n}\). We need to create an expression that simplifies to \(\frac{5x^{7}}{y^{3}}\) using this rule and has at least one negative exponent. Let's start with an expression where we can apply the division rule. Let's consider a numerator with a negative exponent and a denominator, or a denominator with a positive exponent and numerator with a higher exponent. Let's construct the expression: \(\frac{5x^{7}y^{-2}}{y^{1}}\)

Step 2: Apply the division rule

Using \(\frac{a^m}{a^n}=a^{m - n}\) for \(y\) terms: \(\frac{y^{-2}}{y^{1}}=y^{-2-1}=y^{-3}=\frac{1}{y^{3}}\) (by the definition of negative exponents \(a^{-n}=\frac{1}{a^{n}}\))
The \(x\) term \(x^{7}\) and the coefficient \(5\) remain as they are. So \(\frac{5x^{7}y^{-2}}{y^{1}}=\frac{5x^{7}}{y^{3}}\) (since \(y^{-2}\div y^{1}=y^{-3}=\frac{1}{y^{3}}\) and when we multiply by \(5x^{7}\) we get \(\frac{5x^{7}}{y^{3}}\))

Step 1: Recall exponent rules (multiplication and power)

The multiplication rule is \(a^m\cdot a^n=a^{m + n}\) and the power rule is \((a^m)^n=a^{m\times n}\). We need to create an expression that simplifies to \(5^{3}x^{5}y^{2}\) using these rules and has at least two negative exponents. Let's break down the exponents. For \(x^{5}\), we can use power rule: suppose we have \((x^{a})^b=x^{ab}\), let's take \(a = - 1\) and \(b=-5\), then \((x^{-1})^{-5}=x^{(-1)\times(-5)}=x^{5}\). For \(y^{2}\), let's take \((y^{-1})^{-2}=y^{(-1)\times(-2)}=y^{2}\). And for the coefficient \(5^{3}\), we can have a term with negative exponents in the numerator and denominator. Let's construct the expression: \(\frac{5^{3}x^{-1}y^{-1}\cdot(x^{-1})^{-5}\cdot(y^{-1})^{-2}}{x^{0}y^{0}}\) (but we can simplify the construction). Alternatively, let's use multiplication and power rule with negative exponents. Let's create an expression: \(5^{3}x^{-2}y^{-3}\cdot x^{7}y^{5}\)

Step 2: Apply the rules

First, apply the multiplication rule for \(x\) terms: \(x^{-2}\cdot x^{7}=x^{-2 + 7}=x^{5}\)
For \(y\) terms: \(y^{-3}\cdot y^{5}=y^{-3 + 5}=y^{2}\)
The coefficient term \(5^{3}\) remains. Now, let's check the negative exponents: we have \(x^{-2}\) and \(y^{-3}\) which are two negative exponents. Now, let's also use the power rule. Suppose we have \((5^{3})(x^{-2}y^{-3})(x^{7}y^{5})\). Wait, maybe a better construction: Let's use power rule on a term with negative exponents. Let's take \((5x^{-1}y^{-1})^{3}\cdot x^{8}y^{5}\)
First, apply the power rule: \((5x^{-1}y^{-1})^{3}=5^{3}(x^{-1})^{3}(y^{-1})^{3}=5^{3}x^{-3}y^{-3}\) (using \((ab)^n=a^n b^n\) and \((a^m)^n=a^{mn}\))
Then apply the multiplication rule with \(x^{8}y^{5}\):
For \(x\) terms: \(x^{-3}\cdot x^{8}=x^{-3 + 8}=x^{5}\)
For \(y\) terms: \(y^{-3}\cdot y^{5}=y^{-3+5}=y^{2}\)
The coefficient term \(5^{3}\) remains. Now, we have two negative exponents in the initial power - term (\(x^{-3}\) and \(y^{-3}\)). So the expression \((5x^{-1}y^{-1})^{3}\cdot x^{8}y^{5}\) simplifies to \(5^{3}x^{5}y^{2}\)

Step 3: Verify

\((5x^{-1}y^{-1})^{3}=5^{3}(x^{-1})^{3}(y^{-1})^{3}=125x^{-3}y^{-3}\)
Then \(125x^{-3}y^{-3}\cdot x^{8}y^{5}=125x^{-3 + 8}y^{-3 + 5}=125x^{5}y^{2}=5^{3}x^{5}y^{2}\)
And we have two negative exponents (\(x^{-3}\) and \(y^{-3}\) initially from the power - term)

Answer:

\(4^{2}x^{3}y^{2}\cdot x^{4}y^{2}\) (Other valid expressions are also possible, for example \(4^{2}x^{1}y^{1}\cdot x^{6}y^{3}\) etc., as long as the exponents add up correctly using the multiplication rule)

Part b