Problem 75 - Algebra 2 and Trigonometry

Question

In \(64-75,\) write each quotient as a product without a denominator. The variables are not equal to zero.$$ \frac{25 a^{5} b^{-3}}{5^{0} a^{-1} b} $$

Step-by-Step Solution

Answer

The quotient is rewritten as the product: \(25a^6b^{-4}\).

Step 1: Simplify the constants

First, simplify any constants in the expression. Since anything raised to the power of 0 is 1, we simplify \(5^0\) to 1. This gives us \( \frac{25 a^5 b^{-3}}{1 \cdot a^{-1} b} \), which simplifies further to \( \frac{25 a^5 b^{-3}}{a^{-1} b} \).

Step 2: Use laws of exponents

Apply the laws of exponents to combine the powers of \(a\) and \(b\). For division, subtract the exponents: \( a^5 \div a^{-1} = a^{5 - (-1)} = a^{5 + 1} = a^6 \), and \( b^{-3} \div b = b^{-3 - 1} = b^{-4} \). Thus, the expression simplifies to \( 25 a^6 b^{-4} \).

Step 3: Express as a product without denominators

Rewrite the expression by applying the property of negative exponents, which says \( x^{-n} = \frac{1}{x^n} \). So, \( b^{-4} = \frac{1}{b^4} \). We can then express the entire expression as a product without denominators by writing it as \( 25 \cdot a^6 \cdot \frac{1}{b^4} \), which is simply \( 25a^6b^{-4} \).

Key Concepts

Negative Exponents

Negative exponents might look tricky at first, but they're actually pretty simple once you get the hang of them. If you see a negative exponent, it means you flip the base to the bottom of a fraction. For example, if you have something like this:

• \( x^{-n} \)

it becomes:

• \( \frac{1}{x^n} \).

The negative sign in the exponent tells you that you need to take the reciprocal of the base. This is really useful when you're trying to simplify expressions. The original problem involved negative exponents like \( b^{-3} \). By the rules of negative exponents, this is transformed to \( \frac{1}{b^3} \) when looking to simplify or express without a fraction. Keeping this rule in mind makes it much easier when you're converting expressions to something more manageable!

Simplifying Algebraic Expressions

Simplifying algebraic expressions is all about making calculations easier and expressions more understandable. Let's break it down. You want to combine like terms and clear out any fractions or unwanted components.
In this particular exercise, you're given an expression with a mixture of constants and variables in both the numerator and the denominator:

• \( \frac{25 a^{5} b^{-3}}{a^{-1} b} \).

With the constants and variables in place, you first simplify the constants and then apply the exponent rules to the variables.
The denominator included a \(5^0\), but remember, anything raised to the power of zero simplifies to one. This reduces the clutter.
Next, you apply the rules of exponents to combine terms:

• \( a^5 \) and \( a^{-1} \) result in \( a^{6} \) after subtraction of the exponents,
• where \( a^5 \div a^{-1} = a^{5 - (-1)} = a^{5+1} = a^6 \). Similarly,
• for \( b^{-3} \) and \( b \),
• we have: \( b^{-3} \div b = b^{-3-1} = b^{-4}\).

When simplified, you get the expression without any fractions.

Exponent Rules

Mastering exponent rules is essential for simplifying expressions and solving more complex algebraic problems. The

• laws of exponents

include several handy shortcuts:

• The product of powers rule: When you multiply like bases, \( x^m \times x^n \), you add the exponents, so it becomes \(x^{m+n}\).
• The power of a power rule: For an expression \((x^m)^n\), you multiply the exponents to get \(x^{m \cdot n}\).
• The quotient of powers rule: When dividing like bases, you subtract the exponents: \( \frac{x^m}{x^n} = x^{m-n}\).
• Finally, the negative exponent rule: Any \(x^{-n} = \frac{1}{x^n}\), which switches the base from numerator to denominator or vice-versa.

In the solved exercise, the quotient of powers rule was used to simplify terms, with notations like:

• \( a^5 \div a^{-1} \),
• becoming \( a^{5+1} = a^6 \).

Understanding how to use these rules lets you organize and solve expressions in a more structured way, simplifying the complex world of algebra.