### Question 19/26A scientist used the function \( f(m) = 3 \left(2\right)^{2m+1} \) to conduct her experiment. Consider the following expressions:I. \( 6 \left(2\right)^{2m} \) II. \( 6 \left(4\right)^m \) III. \( 6 \left(6\right)^{2m} \)**The function \( f(m) \) is equivalent to:**- **A. I and II**- **B. I, only**- **C. II, only**- **D. I and III****Explanation:**We need to analyze if the given expressions (I, II, and III) are equivalent to the function \( f(m) = 3 \left(2\right)^{2m+1} \).1. **Expression I: \( 6 \left(2\right)^{2m} \)** Rewrite \( f(m) \): \[ f(m) = 3 \cdot 2^{2m} \cdot 2^1 = 3 \cdot 2 \cdot 2^{2m} = 6 \cdot 2^{2m} \] Therefore, \( f(m) \) is equivalent to Expression I.2. **Expression II: \( 6 \left(4\right)^m \)** Rewrite \( f(m) \): \[ f(m) = 3 \cdot 2^{2m+1} = 3 \cdot 2 \cdot 2^{2m} = 6 \cdot 2^{2m} \implies 6 \cdot (2^2)^m = 6 \cdot 4^m \] Therefore, \( f(m) \) is equivalent to Expression II.3. **Expression III: \( 6 \left(6\right)^{2m} \)** Analyze Expression III if: \[ f(m) = 6 \cdot 6^{2m} \] This does not hold as both the bases are different.Thus, \( f(m) \) is equivalent to both Expressions I and II.**Answer: A.

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Answered: 1. 6 (2)²m II. 6 (4) III. 6 (6)2m

Math

Algebra

1. 6 (2)²m II. 6 (4) III. 6 (6)2m

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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