Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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 :

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 this was helpful. Please come back whenever you need more information or answers to your queries. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.