Problem 39
Question
Simplify the variable expression. $$ x^{1 / 3} \cdot x^{1 / 2} $$
Step-by-Step Solution
Verified Answer
\(x^{5/6}\)
1Step 1: Identify the base and the exponents
The base is \(x\) and the exponents are \(1/3\) and \(1/2\). We are multiplying these two expressions together.
2Step 2: Apply the rule of exponentiation
The rule of exponentiation states that when you multiply expressions with the same base, you add the exponents. Therefore, we add \(1/3\) and \(1/2\). This gives us \(x^{1/3+1/2}\).
3Step 3: Add the fractions
To add fractions, you have to find a common denominator. The least common multiple (LCM) of 2 and 3 is 6. Rewrite the fractions as \(\frac{2}{6}\) and \(\frac{3}{6}\), which are equivalent to \(1/3\) and \(1/2\) respectively. Therefore, \(x^{1/3+1/2}\) simplifies to \(x^{2/6 + 3/6}\).
4Step 4: Simplify the expression
Finally, we simplify the expression further by summing up the fractions. This gives us \(x^{5/6}\).
Key Concepts
Rules of ExponentiationAdding Fractions with Different DenominatorsExponents and PowersLeast Common Multiple (LCM)
Rules of Exponentiation
When dealing with variable expressions, understanding the rules of exponentiation can simplify the process immensely. Exponentiation is the operation of raising one quantity, known as the base, to the power of another quantity, known as the exponent. A key rule to remember is that when you multiply expressions with the same base, you simply add their exponents. Consider the expression \(x^{a} \cdot x^{b} = x^{a+b}\). Here, \(x\) is the base and \(a\) and \(b\) are the exponents.
By applying this rule to variable expressions, like \(x^{1/3} \cdot x^{1/2}\), we focus on augmenting the exponents: we add \(1/3\) and \(1/2\) to get a new exponent for the common base \(x\). This straightforward approach streamlines the simplification process and helps prevent common mistakes when manipulating powers and roots.
By applying this rule to variable expressions, like \(x^{1/3} \cdot x^{1/2}\), we focus on augmenting the exponents: we add \(1/3\) and \(1/2\) to get a new exponent for the common base \(x\). This straightforward approach streamlines the simplification process and helps prevent common mistakes when manipulating powers and roots.
Adding Fractions with Different Denominators
Adding fractions can seem daunting when they have different denominators, but there's an easy method to follow — finding a common denominator. This is essential to perform the addition correctly. You can do this by identifying the least common multiple (LCM) of both denominators, which serves as the common denominator. For instance, when adding \(\frac{1}{2}\) and \(\frac{1}{3}\), the denominators are 2 and 3.
Their LCM is 6, so we rewrite each fraction with 6 as the denominator, obtaining \(\frac{3}{6}\) and \(\frac{2}{6}\) respectively. Then, you can easily add the fractions: \(\frac{3}{6} + \frac{2}{6} = \frac{5}{6}\). Through this method, students can confidently combine fractions with different denominators, simplifying expressions without hesitation.
Their LCM is 6, so we rewrite each fraction with 6 as the denominator, obtaining \(\frac{3}{6}\) and \(\frac{2}{6}\) respectively. Then, you can easily add the fractions: \(\frac{3}{6} + \frac{2}{6} = \frac{5}{6}\). Through this method, students can confidently combine fractions with different denominators, simplifying expressions without hesitation.
Exponents and Powers
The words 'exponents' and 'powers' often intermingle in mathematical expressions. To put it plainly, an exponent refers to the number that indicates how many times the base is multiplied by itself. On the other hand, 'power' typically refers to the entire expression, such as \(2^{3}\) being read as 'two to the third power'. Understanding exponents and powers is a fundamental part of algebra. When you encounter a variable to an exponent, like \(x^{n}\), it means \(x\) is being multiplied by itself \(n\) times.
Exponents can also be fractions, which often represent roots, such as \(x^{1/2}\) being equivalent to the square root of \(x\). Grasping how exponents dictate the repeated multiplication can greatly aid in simplifying variable expressions and solving more complex algebraic equations.
Exponents can also be fractions, which often represent roots, such as \(x^{1/2}\) being equivalent to the square root of \(x\). Grasping how exponents dictate the repeated multiplication can greatly aid in simplifying variable expressions and solving more complex algebraic equations.
Least Common Multiple (LCM)
The Least Common Multiple (LCM) is a concept that often comes into play when working with fractions, particularly in adding or subtracting them. The LCM of two or more numbers is the smallest number that is a multiple of each of the numbers. It can be found through various methods, including listing out multiples or using the prime factorization of the numbers.
Finding the LCM helps to identify the smallest common denominator for fractions, which is crucial in combining them into a single fraction. For instance, with the numbers 2 and 3, the smallest number that is a multiple of both is 6. Thus, 6 is their LCM. Knowing how to determine the LCM allows students to easily combine fractions, as shown in simplifying expressions such as \(x^{1/3} \cdot x^{1/2}\), where the common denominator for the fractions \(1/3\) and \(1/2\) is essential in finding the sum of the exponents. By mastering this, mathematical expressions become less complex and students can approach problems with a clear strategy.
Finding the LCM helps to identify the smallest common denominator for fractions, which is crucial in combining them into a single fraction. For instance, with the numbers 2 and 3, the smallest number that is a multiple of both is 6. Thus, 6 is their LCM. Knowing how to determine the LCM allows students to easily combine fractions, as shown in simplifying expressions such as \(x^{1/3} \cdot x^{1/2}\), where the common denominator for the fractions \(1/3\) and \(1/2\) is essential in finding the sum of the exponents. By mastering this, mathematical expressions become less complex and students can approach problems with a clear strategy.
Other exercises in this chapter
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