Question
(c) \( 2^{\log _{2 \sqrt{2}} 27} \cdot L \)
Ask by Montgomery Henry. in Bangladesh
Mar 06,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The simplified expression is \( 9L \).
Solution
To simplify the expression \( 2^{\log_{2 \sqrt{2}} 27} \cdot L \), we can start by simplifying the logarithmic part.
1. **Change of Base Formula**: We can use the change of base formula for logarithms, which states that:
\[
\log_a b = \frac{\log_c b}{\log_c a}
\]
In this case, we will change the base from \( 2 \sqrt{2} \) to \( 2 \).
2. **Calculate \( \log_{2 \sqrt{2}} 27 \)**:
\[
\log_{2 \sqrt{2}} 27 = \frac{\log_2 27}{\log_2 (2 \sqrt{2})}
\]
3. **Calculate \( \log_2 (2 \sqrt{2}) \)**:
\[
\log_2 (2 \sqrt{2}) = \log_2 2 + \log_2 \sqrt{2} = 1 + \frac{1}{2} = \frac{3}{2}
\]
4. **Calculate \( \log_2 27 \)**:
\[
\log_2 27 = \log_2 (3^3) = 3 \log_2 3
\]
5. **Substituting back**:
\[
\log_{2 \sqrt{2}} 27 = \frac{3 \log_2 3}{\frac{3}{2}} = 2 \log_2 3
\]
6. **Substituting into the original expression**:
\[
2^{\log_{2 \sqrt{2}} 27} = 2^{2 \log_2 3}
\]
7. **Using the property of exponents**:
\[
2^{2 \log_2 3} = (2^{\log_2 3})^2 = 3^2 = 9
\]
8. **Final expression**:
\[
2^{\log_{2 \sqrt{2}} 27} \cdot L = 9 \cdot L
\]
Thus, the simplified expression is:
\[
9L
\]
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Simplify this solution Extra Insights
To simplify the expression \( 2^{\log _{2 \sqrt{2}} 27} \cdot L \), we first focus on the logarithm. Recall that \( \log_{a^b} c = \frac{1}{b} \log_a c \). Here, since \( 2 \sqrt{2} = 2^{1.5} \), we can write the logarithm as: \[ \log_{2 \sqrt{2}} 27 = \frac{1}{1.5} \log_2 27 = \frac{2}{3} \log_2 27. \] Now substituting this into our expression gives us: \[ 2^{\log_{2 \sqrt{2}} 27} = 2^{\frac{2}{3} \log_2 27} = 27^{\frac{2}{3}}. \] This means that \( 2^{\log_{2 \sqrt{2}} 27} = \sqrt[3]{27^2} = \sqrt[3]{729} = 9 \). Thus, the entire expression simplifies to: \[ 9 \cdot L. \]
