Problem 53 - Algebra 2

Question

Simplify each number. $$243^{1.2}$$

Step-by-Step Solution

Answer

The simplified form of \(243^{1.2}\) is 729.

Step 1: Identify the base and the exponent

The given number is \(243^{1.2}\). Here, 243 is our base, and 1.2 is our exponent. The exponent implies that we are multiplying 243 by itself 1.2 times, which involves a combination of multiplication and taking roots.

Step 2: Convert the decimal exponent into fraction

It's easier to handle the exponent when it's in fraction form. The decimal 1.2 can be written as \( \frac{6}{5} \). So, the number becomes \(243^{6/5}\).

Step 3: Split the exponentiation into a root and a power

This step transforms the number into a multiplication operation and a fifth root operation. The denominator of our fraction shows the root we have to take, and the numerator gives the power. Hence, we rewrite \(243^{6/5}\) as \( \sqrt[5]{243^6}\).

Step 4: Simplify the number

We know that the fifth root of 243 is 3 (since \(3^5 = 243\)). So we rewrite \( \sqrt[5]{243^6}\) as \( 3^6\). When calculated, \( 3^6\) equals to 729.

Key Concepts

Decimal to Fraction Conversion

When dealing with powers and exponents, especially those in decimal form, quickly converting them into fractions can make calculations much more manageable. Let's start by understanding how this conversion works.
A decimal number such as 1.2 can be expressed as a fraction for easier computation. The idea is to recognize the place value of the decimal. For 1.2, the decimal part '2' is in the tenths place, which means it represents 2 tenths or \( \frac{2}{10} \).
To further simplify, divide the numerator and the denominator by their greatest common divisor, which is 2 in this case. Thus, \( \frac{2}{10} \) simplifies to \( \frac{1}{5} \).
Hence, 1.2 becomes \( \frac{6}{5} \) when considered as a mixed number (1 whole + \( \frac{2}{5} \) \( = \frac{5}{5} + \frac{1}{5} \)). This understanding transforms calculations, making operations with exponents far simpler.

Fractional Exponents

Fractional exponents might seem complex, but they are essentially another way to represent roots combined with powers. To decode fractional exponents, remember the form \( a^{m/n} \). Here, you can interpret \( m \) as the power and \( n \) as the root.
Consider \( 243^{6/5} \), where \( 6 \) is the power and \( 5 \) is the root. This means that you'll be raising 243 to the sixth power, after which you'll take a fifth root.
Understanding this concept enables you to represent and solve expressions involving roots and exponents systematically. Moreover, fractional exponents offer a streamlined method to avoid intricate root operations, allowing simplification into usable forms for further calculations.

Root and Power Transformation

Once you've represented your problem with fractional exponents, the next step is to separate the exponent into two operations: a root and then a power.
Taking the problem \( 243^{6/5} \), we first focus on the root by observing the denominator, 5. This indicates a fifth root operation. The transformation looks something like \( \sqrt[5]{243^6} \). This change breaks the complex operation into more manageable steps.
The root and power transformation splits the calculation into digestible parts, making it simpler to solve large exponent problems progressively. In this instance, finding the fifth root first helps immensely, simplifying the expression significantly before dealing with the power.

Exponentiation

Exponentiation allows numbers to be expressed as repeated multiplications. It emerges in various forms, including fractional, as mentioned.
With \( 243^{1.2} \) known now as \( \sqrt[5]{243^6} \), understanding exponentiation focuses on calculating \( 3^6 \) after simplifying with a root. Knowing the fifth root of 243 is 3 (derived from \( 3^5 = 243 \)), exponentiation turns the expression into something more straightforward: \( 3^6 \).
Exponentiation itself is straightforward once familiar with handling powers and roots. It acts as a powerful tool in mathematics, allowing complex expressions to be managed in smaller, calculable sections, which greatly simplifies arithmetic operations, from algebra to calculus.