What is $\frac{\left(x^2 y^3\right)^{\frac{1}{3}}}{\sqrt[3]{x^2 y}}$ in exponential form?
Answer (2)
The expression 3 x 2 y ( x 2 y 3 ) 3 1 simplifies to y 3 2 in exponential form. This was achieved by simplifying both the numerator and denominator using exponent rules. Ultimately, the final result is y 3 2 .
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Simplify the numerator: ( x 2 y 3 ) 3 1 = x 3 2 y .
Simplify the denominator: 3 x 2 y = x 3 2 y 3 1 .
Divide the numerator by the denominator: x 3 2 y 3 1 x 3 2 y = y 3 2 .
The correct answer is x 3 2 y 3 1 x 3 2 y .
Explanation
Understanding the Problem We are given the expression 3 x 2 y ( x 2 y 3 ) 3 1 . We need to simplify this expression and express it in exponential form.
Simplifying the Numerator First, let's simplify the numerator: ( x 2 y 3 ) 3 1 = x 3 2 y 3 3 = x 3 2 y
Simplifying the Denominator Next, let's simplify the denominator: 3 x 2 y = ( x 2 y ) 3 1 = x 3 2 y 3 1
Dividing Numerator by Denominator Now, divide the numerator by the denominator: x 3 2 y 3 1 x 3 2 y = x 3 2 − 3 2 y 1 − 3 1 = x 0 y 3 2 = y 3 2
Comparing with Given Options Now, let's examine the given options to see which one matches our simplified expression, y 3 2 .
Option 1: x 3 2 y 3 1 x 3 2 y = x 3 2 − 3 2 y 1 − 3 1 = x 0 y 3 2 = y 3 2 . This matches our simplified expression.
Option 2: x 3 7 y 3 2 x 3 7 y 3 10 = x 3 7 − 3 7 y 3 10 − 3 2 = x 0 y 3 8 = y 3 8 . This does not match our simplified expression.
Option 3: x 5 5 y 5 2 x 3 5 y 3 9 = x 1 y 5 2 x 3 5 y 3 = x 3 5 − 1 y 3 − 5 2 = x 3 2 y 5 13 . This does not match our simplified expression.
Final Answer Therefore, the correct answer is Option 1: x 3 2 y 3 1 x 3 2 y
Examples
Understanding exponential forms and simplification is crucial in various fields, such as physics and engineering, where complex equations are frequently encountered. For instance, when dealing with wave functions in quantum mechanics, simplifying expressions involving exponents and radicals is essential for solving the Schrödinger equation and predicting the behavior of particles. Similarly, in electrical engineering, simplifying expressions with exponents is necessary when analyzing circuits and calculating power dissipation. Mastering these simplification techniques allows for more efficient problem-solving and a deeper understanding of the underlying principles.
