We appreciate your visit to Identify the expression equivalent to tex frac log 2 128 log 2 16 tex A tex log 16 128 tex B tex log 2 128. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
To solve the problem, we want to find which of the given expressions is equivalent to [tex]\(\frac{\log _2 128}{\log _2 16}\)[/tex].
### Step-by-step Solution:
1. Understand the Given Expression:
- The expression is [tex]\(\frac{\log _2 128}{\log _2 16}\)[/tex].
- This can be interpreted as "the logarithm of 128 with base 2 divided by the logarithm of 16 with base 2".
2. Logical Simplification:
- Both the numerator and the denominator are logarithms with the same base (base 2).
- Using the property of logarithms, [tex]\(\frac{\log_b a}{\log_b c} = \log_c a\)[/tex], we can simplify this expression directly to [tex]\(\log_{16} 128\)[/tex].
3. Evaluate Each Option:
- [tex]\(\log_{16} 128\)[/tex]: Based on the simplification above, [tex]\(\frac{\log _2 128}{\log _2 16}\)[/tex] equals [tex]\(\log_{16} 128\)[/tex].
- [tex]\(\log_2 128\)[/tex], [tex]\(\log_4 128\)[/tex], and [tex]\(\log_{128} 16\)[/tex] are not equivalent to the simplified expression based on the logarithmic properties we've used.
Therefore, the expression [tex]\(\frac{\log _2 128}{\log _2 16}\)[/tex] is equivalent to [tex]\(\log_{16} 128\)[/tex].
### Step-by-step Solution:
1. Understand the Given Expression:
- The expression is [tex]\(\frac{\log _2 128}{\log _2 16}\)[/tex].
- This can be interpreted as "the logarithm of 128 with base 2 divided by the logarithm of 16 with base 2".
2. Logical Simplification:
- Both the numerator and the denominator are logarithms with the same base (base 2).
- Using the property of logarithms, [tex]\(\frac{\log_b a}{\log_b c} = \log_c a\)[/tex], we can simplify this expression directly to [tex]\(\log_{16} 128\)[/tex].
3. Evaluate Each Option:
- [tex]\(\log_{16} 128\)[/tex]: Based on the simplification above, [tex]\(\frac{\log _2 128}{\log _2 16}\)[/tex] equals [tex]\(\log_{16} 128\)[/tex].
- [tex]\(\log_2 128\)[/tex], [tex]\(\log_4 128\)[/tex], and [tex]\(\log_{128} 16\)[/tex] are not equivalent to the simplified expression based on the logarithmic properties we've used.
Therefore, the expression [tex]\(\frac{\log _2 128}{\log _2 16}\)[/tex] is equivalent to [tex]\(\log_{16} 128\)[/tex].
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