Use the values $$ \log 128 \approx 2.1 $$ and $$ \log 4 \approx 0.6 $$ to find the approximate value of $$ \log _{4} 128 $$. $$ \log _{4} 128 \approx 6 $$ The solution is

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The approximate value of $$ \log_{4} 128 $$ is 3.5. To find $$ \log_{4} 128 $$, we can use the change of base formula for logarithms, which states: $$ \log_{b} a = \frac{\log_{k} a}{\log_{k} b} $$ where $$ k $$ can be any positive number. In this case, we can use base 10 logarithms (common logarithms) for our calculations. Thus, we have: $$ \log_{4} 128 = \frac{\log 128}{\log 4} $$ Given the approximate values: - $$ \log 128 \approx 2.1 $$ - $$ \log 4 \approx 0.6 $$ We can substitute these values into the formula: $$ \log_{4} 128 \approx \frac{2.1}{0.6} $$ Now, we perform the division: $$ \frac{2.1}{0.6} = \frac{21}{6} = 3.5 $$ Thus, the approximate value of $$ \log_{4} 128 $$ is: $$ \log_{4} 128 \approx 3.5 $$ So the solution is: $$ \log_{4} 128 \approx 3.5 $$

Use the values \( \log 128 \approx 2.1 \) and \( \log 4 \approx 0.6 \) to find the approximate value of \( \log _{4} 128 \). \( \log _{4} 128 \approx 6 \) The solution is

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Use the values \( \log 128 \approx 2.1 \) and \( \log 4 \approx 0.6 \) to find the approximate value of \( \log _{4} 128 \). \( \log _{4} 128 \approx 6 \) The solution is

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