126. $$ \log _{8} 143+\log _{9} 96 $$

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The simplified expression is approximately 4.48. To simplify the expression $$ \log_{8} 143 + \log_{9} 96 $$, we can use the change of base formula for logarithms, which states: $$ \log_{b} a = \frac{\log_{k} a}{\log_{k} b} $$ for any positive $$ k $$. We will use base 10 for convenience. First, we convert $$ \log_{8} 143 $$: $$ \log_{8} 143 = \frac{\log_{10} 143}{\log_{10} 8} $$ Next, we convert $$ \log_{9} 96 $$: $$ \log_{9} 96 = \frac{\log_{10} 96}{\log_{10} 9} $$ Now, substituting these into the original expression gives: $$ \log_{8} 143 + \log_{9} 96 = \frac{\log_{10} 143}{\log_{10} 8} + \frac{\log_{10} 96}{\log_{10} 9} $$ To combine these fractions, we need a common denominator, which is $$ \log_{10} 8 \cdot \log_{10} 9 $$: $$ \log_{8} 143 + \log_{9} 96 = \frac{\log_{10} 143 \cdot \log_{10} 9 + \log_{10} 96 \cdot \log_{10} 8}{\log_{10} 8 \cdot \log_{10} 9} $$ Next, we can calculate the logarithms using a calculator or logarithm tables: - $$ \log_{10} 143 \approx 2.155 $$ - $$ \log_{10} 96 \approx 1.982 $$ - $$ \log_{10} 8 = \log_{10} (2^3) = 3 \log_{10} 2 \approx 3 \times 0.301 = 0.903 $$ - $$ \log_{10} 9 = \log_{10} (3^2) = 2 \log_{10} 3 \approx 2 \times 0.477 = 0.954 $$ Now substituting these values into our expression: $$ \log_{8} 143 + \log_{9} 96 \approx \frac{(2.155)(0.954) + (1.982)(0.903)}{(0.903)(0.954)} $$ Calculating the numerator: $$ (2.155)(0.954) \approx 2.056 + (1.982)(0.903) \approx 1.794 $$ So, $$ 2.056 + 1.794 \approx 3.850 $$ Now calculating the denominator: $$ (0.903)(0.954) \approx 0.859 $$ Finally, we compute: $$ \log_{8} 143 + \log_{9} 96 \approx \frac{3.850}{0.859} \approx 4.48 $$ Thus, the final answer is: $$ \boxed{4.48} $$

126. \( \log _{8} 143+\log _{9} 96 \)

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126. \( \log _{8} 143+\log _{9} 96 \)

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