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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]
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