To solve for \( x \) in the equation:
\[ \log_b x = \frac{2}{3} \log_b 27 + 2 \log_b 2 - \log_b 3, \]
we can use the properties of logarithms to simplify and solve for \( x \).
Step-by-Step Solution:
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Apply the Power Rule:
The power rule of logarithms states that \( \log_b a^n = n \log_b a \). Apply this to each term:
\[ \log_b x = \frac{2}{3} \log_b 27 + 2 \log_b 2 - \log_b 3 \]
This can be rewritten as:
\[ \log_b x = \log_b 27^{\frac{2}{3}} + \log_b 2^2 - \log_b 3 \]
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Simplify the Exponents:
Calculate \( 27^{\frac{2}{3}} \) and \( 2^2 \):
\[ 27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{3 \times \frac{2}{3}} = 3^2 = 9 \]
\[ 2^2 = 4 \]
So the equation becomes:
\[ \log_b x = \log_b 9 + \log_b 4 - \log_b 3 \]
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Combine the Logarithms:
Use the product and quotient rules of logarithms:
\[ \log_b x = \log_b (9 \times 4) - \log_b 3 = \log_b 36 - \log_b 3 \]
\[ \log_b x = \log_b \left(\frac{36}{3}\right) = \log_b 12 \]
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Solve for \( x \):
Since \( \log_b x = \log_b 12 \), it follows that:
\[ x = 12 \]
Answer:
\[ \boxed{x = 12} \]
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