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Solve log(6) 9-log(9) 27 + log(8)x = log...

Solve `log_(6) 9-log_(9) 27 + log_(8)x = log_(64) x - log_(6) 4`..

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To solve the equation \( \log_{6} 9 - \log_{9} 27 + \log_{8} x = \log_{64} x - \log_{6} 4 \), we will follow these steps: ### Step 1: Rewrite the logarithmic expressions Using the properties of logarithms, we can rewrite the equation: \[ \log_{6} 9 - \log_{9} 27 + \log_{8} x = \log_{64} x - \log_{6} 4 \] We can use the change of base formula and properties of logarithms to simplify the terms. ### Step 2: Simplify \( \log_{9} 27 \) We know that: \[ \log_{9} 27 = \frac{\log_{6} 27}{\log_{6} 9} \] Since \( 27 = 3^3 \) and \( 9 = 3^2 \), we can express: \[ \log_{6} 27 = 3 \log_{6} 3 \quad \text{and} \quad \log_{6} 9 = 2 \log_{6} 3 \] Thus, \[ \log_{9} 27 = \frac{3 \log_{6} 3}{2 \log_{6} 3} = \frac{3}{2} \] ### Step 3: Substitute back into the equation Now substitute \( \log_{9} 27 \) back into the equation: \[ \log_{6} 9 - \frac{3}{2} + \log_{8} x = \log_{64} x - \log_{6} 4 \] ### Step 4: Rewrite \( \log_{64} x \) We can also express \( \log_{64} x \): \[ \log_{64} x = \frac{\log_{8} x}{\log_{8} 64} = \frac{\log_{8} x}{2} \] since \( 64 = 8^2 \). ### Step 5: Substitute and simplify Now, substituting this into the equation gives: \[ \log_{6} 9 - \frac{3}{2} + \log_{8} x = \frac{1}{2} \log_{8} x - \log_{6} 4 \] ### Step 6: Combine like terms Rearranging gives: \[ \log_{6} 9 + \log_{6} 4 - \frac{3}{2} = -\frac{1}{2} \log_{8} x + \log_{8} x \] This simplifies to: \[ \log_{6} 9 + \log_{6} 4 - \frac{3}{2} = \frac{1}{2} \log_{8} x \] ### Step 7: Solve for \( \log_{8} x \) Multiply both sides by 2: \[ 2(\log_{6} 9 + \log_{6} 4 - \frac{3}{2}) = \log_{8} x \] This simplifies to: \[ 2 \log_{6} 9 + 2 \log_{6} 4 - 3 = \log_{8} x \] ### Step 8: Rewrite in exponential form Now we can rewrite this in exponential form: \[ x = 8^{(2 \log_{6} 9 + 2 \log_{6} 4 - 3)} \] ### Step 9: Calculate the value of \( x \) Using the properties of logarithms, we can evaluate: \[ x = 8^{(2 \log_{6} 36 - 3)} = 8^{(2 \cdot 1 - 3)} = 8^{-1} = \frac{1}{8} \] ### Final Answer Thus, the solution is: \[ \boxed{\frac{1}{8}} \]

To solve the equation \( \log_{6} 9 - \log_{9} 27 + \log_{8} x = \log_{64} x - \log_{6} 4 \), we will follow these steps: ### Step 1: Rewrite the logarithmic expressions Using the properties of logarithms, we can rewrite the equation: \[ \log_{6} 9 - \log_{9} 27 + \log_{8} x = \log_{64} x - \log_{6} 4 \] We can use the change of base formula and properties of logarithms to simplify the terms. ...
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