Problem 204 - Intermediate Algebra

Question

In the following exercises, simplify. $$ \frac{\left(\frac{3}{4}\right)^{2}}{\left(\frac{5}{8}\right)^{2}} $$

Step-by-Step Solution

Answer

The simplified form is \( \frac{36}{25} \).

Step 1: Understand the Problem

The task is to simplify the given expression \( \frac{\big( \frac{3}{4} \big)^{2}}{\big( \frac{5}{8} \big)^{2}} \). This expression involves fractions within fractions that are squared.

Step 2: Apply the Power Rule for Fractions

For any fraction \( \frac{a}{b} \) raised to a power, the rule \[ \bigg( \frac{a}{b} \bigg)^{n} = \frac{a^{n}}{b^{n}} \] applies. Hence, simplify both the numerator and the denominator separately using this rule: \[ \frac{\big( \frac{3}{4} \big)^{2}}{\big( \frac{5}{8} \big)^{2}} = \frac{\frac{3^{2}}{4^{2}}}{\frac{5^{2}}{8^{2}}} \].

Step 3: Simplify Numerator and Denominator Separately

Calculate the numerator and the denominator: \[ \frac{3^{2}}{4^{2}} = \frac{9}{16} \] and \[ \frac{5^{2}}{8^{2}} = \frac{25}{64} \]. Therefore, the expression becomes \[ \frac{\frac{9}{16}}{\frac{25}{64}} \].

Step 4: Simplify the Complex Fraction

Simplify \( \frac{\frac{9}{16}}{\frac{25}{64}} \) by multiplying by the reciprocal of the denominator: \[ \frac{9}{16} \times \frac{64}{25} \].

Step 5: Multiply the Fractions

Perform the multiplication: \[ \frac{9 \times 64}{16 \times 25} \].

Step 6: Simplify the Result

Simplify the result of the multiplication. Since \( 64 \text{ and } 16 \text{ have a common factor} (16) \), and similarly \( 9 \text{ and } 25 \text{ do not have a common factor} \): \[ \frac{9 \times 64}{16 \times 25} = \frac{9 \times 4}{25} = \frac{36}{25} \].

Key Concepts

fraction arithmetic

When working with fractions, understanding basic fraction arithmetic is key. Fractions consist of a numerator (the top part) and a denominator (the bottom part). To simplify complex fractions, like those in the given exercise, you must follow several steps:

First, ensure all fractions are in their simplest form. For example, if you had \(\frac{6}{12}\), you would reduce it to \(\frac{1}{2}\) by dividing both the numerator and the denominator by their greatest common divisor.

Next, when dividing one fraction by another, like in the problem \(\frac{\frac{9}{16}}{\frac{25}{64}}\), you multiply by the reciprocal of the second fraction. The reciprocal of \(\frac{25}{64}\) is \(\frac{64}{25}\). This transforms the division into a multiplication problem: \(\frac{9}{16} \times \frac{64}{25}\).

When multiplying fractions, multiply the numerators together and the denominators together: \(\frac{9 \times 64}{16 \times 25}\).

Finally, always simplify the result. In this case, the result is \(\frac{36}{25}\), already in its simplest form.

exponents

Exponents are a way to express repeated multiplication of a number by itself. For instance, \(3^2\) means \(3 \times 3 = 9\).

When dealing with fractions, the exponent applies both to the numerator and the denominator. For example, \(\big( \frac{3}{4} \big)^2\) means \(\frac{3^2}{4^2}\) which simplifies to \(\frac{9}{16}\).

In the given exercise, each fraction in the numerator and the denominator is squared:
- Numerator: \(\frac{3^2}{4^2} = \frac{9}{16}\)
- Denominator: \(\frac{5^2}{8^2} = \frac{25}{64}\).

This step transforms the initial complex fraction into a simpler one, where the power rule makes the arithmetic straightforward and manageable.

power rule

The power rule in mathematics states that when raising a fraction to a power, you apply the power to both the numerator and the denominator separately. Mathematically, this is expressed as:

\[\bigg( \frac{a}{b} \bigg)^n = \frac{a^n}{b^n}\]

In our example, apply the power rule to both parts of the given complex fraction:

Numerator: \[\bigg( \frac{3}{4} \bigg)^2 = \frac{3^2}{4^2} = \frac{9}{16}\]

Denominator: \[\bigg( \frac{5}{8} \bigg)^2 = \frac{5^2}{8^2} = \frac{25}{64}\].

Now, the complex fraction becomes \[\frac{\frac{9}{16}}{\frac{25}{64}}\]. The power rule simplifies the calculation, allowing you to more easily handle fractions with exponents.