Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Select all the correct answers.

Which expressions are equivalent to [tex]\log _4\left(\frac{1}{4} x^2\right)[/tex]?

A. [tex]\log _4\left(\frac{1}{4}\right)+\log _4 x^2[/tex]

B. [tex]2 \log _4\left(\frac{1}{4}\right)-\log _4 x^2[/tex]

C. [tex]-2+2 \log _4 x[/tex]

D. [tex]-1+2 \log _4 x[/tex]

E. [tex]2 \log _4\left(\frac{1}{4} x\right)[/tex]

Sagot :

GinnyAnswer
To determine which expressions are equivalent to \(\log_4\left(\frac{1}{4} x^2\right)\), we will use logarithm properties, such as the product rule \(\log_b(xy) = \log_b(x) + \log_b(y)\) and the power rule \(\log_b(x^n) = n \log_b(x)\).

1. \(\log_4\left(\frac{1}{4} x^2\right)\):
- Using the product rule for logarithms: \(\log_4\left(\frac{1}{4} x^2\right) = \log_4\left(\frac{1}{4}\right) + \log_4(x^2)\).
- Using the power rule for logarithms: \(\log_4(x^2) = 2 \log_4(x)\).
- Therefore, \(\log_4\left(\frac{1}{4} x^2\right) = \log_4\left(\frac{1}{4}\right) + 2 \log_4(x)\).

Now let's analyze each of the given expressions:

1. \(\log_4\left(\frac{1}{4}\right) + \log_4 x^2\):
- Simplifying the term on the right: \(\log_4 x^2 = 2 \log_4(x)\).
- Therefore, \(\log_4\left(\frac{1}{4}\right) + \log_4 x^2 = \log_4\left(\frac{1}{4}\right) + 2 \log_4(x)\), which is equivalent to the original expression.

2. \(2 \log_4\left(\frac{1}{4}\right) - \log_4 x^2\):
- Simplifying the terms: \(2 \log_4\left(\frac{1}{4}\right) = 2 (-1) = -2\) since \(\log_4(4^{-1}) = -1\).
- \(\log_4 x^2 = 2 \log_4(x)\).
- So, \(2 \log_4\left(\frac{1}{4}\right) - \log_4 x^2 = -2 - 2 \log_4(x)\), which is not equivalent to the original expression.

3. \(-2 + 2 \log_4 x\):
- This form does not equate to the original expression because \(\log_4\left(\frac{1}{4} x^2\right) = \log_4\left(\frac{1}{4}\right) + 2 \log_4(x)\), where \(\log_4\left(\frac{1}{4}\right) = -1\).
- The correct expression has \(-1\) not \(-2\).

4. \(-1 + 2 \log_4 x\):
- Since \(\log_4\left(\frac{1}{4}\right) = -1\) and \(\log_4 x^2 = 2 \log_4(x)\), this form \(-1 + 2 \log_4 x\) is equivalent to the original expression.

5. \(2 \log_4\left(\frac{1}{4} x\right)\):
- Using the product rule: \(\log_4\left(\frac{1}{4} x\right) = \log_4\left(\frac{1}{4}\right) + \log_4(x)\).
- So \(2 \log_4\left(\frac{1}{4} x\right) = 2 \left(\log_4\left(\frac{1}{4}\right) + \log_4(x)\right) = 2(-1 + \log_4(x)) = -2 + 2 \log_4(x)\).
- This form \(-2 + 2 \log_4(x)\) is not equivalent to the original expression.

Based on these steps, the expressions equivalent to \(\log_4\left(\frac{1}{4} x^2\right)\) are:
- \(\log_4\left(\frac{1}{4}\right) + \log_4 x^2\)
- [tex]\(-1 + 2 \log_4 x\)[/tex]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.