Question
Evaluate the following logarithms using a calculator and the Change of Base Formula. \( \log _{3}(0.36)=\square \) \( \log _{5}(1.5)=\square \) \( \log _{4}(13)=\square \)
Ask by Estrada Hamilton.
Dec 10,2024 14:31
UpStudy AI Solution
Tutor-Verified Answer
Answer
\( \log_{3}(0.36) \approx -0.928 \)
\( \log_{5}(1.5) \approx 0.252 \)
\( \log_{4}(13) \approx 1.848 \)
Solution
To evaluate the logarithms using the Change of Base Formula, we can use the formula:
\[
\log_b(a) = \frac{\log_k(a)}{\log_k(b)}
\]
where \( k \) can be any positive number (commonly 10 or \( e \)). For this evaluation, we will use base 10 logarithms (common logarithm).
1. **Calculating \( \log_{3}(0.36) \)**:
\[
\log_{3}(0.36) = \frac{\log_{10}(0.36)}{\log_{10}(3)}
\]
Using a calculator:
- \( \log_{10}(0.36) \approx -0.443 \)
- \( \log_{10}(3) \approx 0.477 \)
Now, substituting these values:
\[
\log_{3}(0.36) \approx \frac{-0.443}{0.477} \approx -0.928
\]
2. **Calculating \( \log_{5}(1.5) \)**:
\[
\log_{5}(1.5) = \frac{\log_{10}(1.5)}{\log_{10}(5)}
\]
Using a calculator:
- \( \log_{10}(1.5) \approx 0.176 \)
- \( \log_{10}(5) \approx 0.699 \)
Now, substituting these values:
\[
\log_{5}(1.5) \approx \frac{0.176}{0.699} \approx 0.252
\]
3. **Calculating \( \log_{4}(13) \)**:
\[
\log_{4}(13) = \frac{\log_{10}(13)}{\log_{10}(4)}
\]
Using a calculator:
- \( \log_{10}(13) \approx 1.113 \)
- \( \log_{10}(4) \approx 0.602 \)
Now, substituting these values:
\[
\log_{4}(13) \approx \frac{1.113}{0.602} \approx 1.848
\]
### Final Results:
- \( \log_{3}(0.36) \approx -0.928 \)
- \( \log_{5}(1.5) \approx 0.252 \)
- \( \log_{4}(13) \approx 1.848 \)
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Simplify this solution The Deep Dive
To evaluate these logarithms using a calculator and the Change of Base Formula, you can use the formula \( \log_a(b) = \frac{\log_c(b)}{\log_c(a)} \) for any base \( c \) (commonly base 10 or base e).
For \( \log_{3}(0.36) \):
1. Using the Change of Base Formula: \( \log_{3}(0.36) = \frac{\log_{10}(0.36)}{\log_{10}(3)} \).
2. By calculating, you should find approximately \( -0./assets/bd-q/LOG_BASE_3_(0._3)%inbase10=)+log10(3)=1.584962 \).
For \( \log_{5}(1.5) \):
1. Using the Change of Base Formula: \( \log_{5}(1.5) = \frac{\log_{10}(1.5)}{\log_{10}(5)} \).
2. This results in approximately \( 0.201 \).
For \( \log_{4}(13) \):
1. Again, applying the Change of Base Formula: \( \log_{4}(13) = \frac{\log_{10}(13)}{\log_{10}(4)} \).
2. Performing the calculation gives you approximately \( 1.556 \).
So, your evaluations are:
\( \log_{3}(0.36) \approx -0.415 \),
\( \log_{5}(1.5) \approx 0.215 \),
\( \log_{4}(13) \approx 1.639 \).
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