What is the following product?
[tex]$\sqrt[3]{5} \cdot \sqrt{2}$[/tex]
A. [tex]$\sqrt[6]{10}$[/tex]
B. [tex]$\sqrt[6]{200}$[/tex]
C. [tex]$\sqrt[6]{500}$[/tex]
D. [tex]$\sqrt[6]{100000}$[/tex]
Answer (2)
The product 3 5 ⋅ 2 simplifies to 6 200 . Therefore, the correct choice is option B: 6 200 .
;
Rewrite the expression using fractional exponents: 3 5 ⋅ 2 = 5 3 1 ⋅ 2 2 1 .
Find a common denominator for the exponents: 5 3 1 ⋅ 2 2 1 = 5 6 2 ⋅ 2 6 3 .
Simplify the exponents: 5 6 2 ⋅ 2 6 3 = ( 5 2 ) 6 1 ⋅ ( 2 3 ) 6 1 = ( 25 ) 6 1 ⋅ ( 8 ) 6 1 .
Combine the terms under a single sixth root: ( 25 ) 6 1 ⋅ ( 8 ) 6 1 = ( 25 ⋅ 8 ) 6 1 = ( 200 ) 6 1 = 6 200 .
6 200
Explanation
Understanding the Problem We are given the expression 3 5 ⋅ 2 and five options: 6 10 , 6 200 , 6 500 , 6 100000 . Our goal is to determine which of the five options is equal to the given expression.
Rewriting with Fractional Exponents Let's rewrite 3 5 and 2 using fractional exponents. Recall that n x = x n 1 . Therefore, we have 3 5 = 5 3 1 and 2 = 2 2 1 .
Finding a Common Denominator Now, let's rewrite the expression 3 5 ⋅ 2 as 5 3 1 ⋅ 2 2 1 . To compare this with the given options, we need to find a common denominator for the exponents. The least common denominator of 3 and 2 is 6. So, we rewrite the exponents with a denominator of 6: 5 3 1 = 5 6 2 and 2 2 1 = 2 6 3 .
Simplifying the Exponents We can now rewrite the expression as 5 6 2 ⋅ 2 6 3 . Using the property x b a = ( x a ) b 1 , we have 5 6 2 = ( 5 2 ) 6 1 and 2 6 3 = ( 2 3 ) 6 1 . This simplifies to ( 25 ) 6 1 ⋅ ( 8 ) 6 1 .
Combining Under a Single Root Now, we can combine the terms under a single sixth root. Recall that x n 1 ⋅ y n 1 = ( x ⋅ y ) n 1 . Therefore, ( 25 ) 6 1 ⋅ ( 8 ) 6 1 = ( 25 ⋅ 8 ) 6 1 = ( 200 ) 6 1 . This is equal to 6 200 .
Final Answer Comparing our result 6 200 with the given options, we see that it matches the second option. Therefore, the product 3 5 ⋅ 2 is equal to 6 200 .
Conclusion Therefore, the correct answer is 6 200 .
Examples
Understanding roots and exponents is crucial in various fields, such as engineering and physics. For instance, when calculating the period of a pendulum, the formula involves a square root. Similarly, understanding exponents is essential in calculating compound interest in finance. By simplifying expressions with roots and exponents, we can solve complex problems in these areas more efficiently. This skill is also valuable in computer graphics, where scaling and transformations often involve manipulating exponents and roots.
