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This could be a basic question. But I would like to know steps I should follow to calculate $\log_2\, 0.667$.


In an answer I found it says $(0.038 \log_2 0.038) = -0.181$. How this calculation works?

Is it $0.038 \log(0.038) / \log 2$ ?

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What tools do you have? On a calculator something like log(0.667)/log(2) would give you the answer, and it does not even matter whether your calculator uses common or natural logarithms. – Henry Mar 5 '12 at 14:23
If you want a fully elementary solution, you can always use the expansion of $\log_2$ as a power series... – Jakub Konieczny Mar 5 '12 at 14:27
Thank you for your suggestions. – Prasad Rajapaksha Mar 5 '12 at 14:31
this may help. – David Mitra Mar 5 '12 at 14:36
You are correct. Using the change-of-base formula, $\log_2(0.038)$ can be written as $\frac{\log(0.038)}{\log(2)}$. – josh Mar 5 '12 at 15:03
up vote 5 down vote accepted

For a rough approximation notice that

$0.667 \approx 2/3$

so $\log_2\left( 2/3\right) = \log_2\left( 2\right) - \log_2\left( 3\right) = 1 - \log_2\left( 3\right)$

From here one can use a change of bases. Like in the answers above

$ \log 0.667 \approx 1 - \frac{\ln 3}{\ln 2}$

(For the wise kids we know that $\ln 2 \approx 0.69$ and $\ln 3 \approx 1.1$ so )

$ \log 0.667 \approx 1 - \frac{\ln 3}{\ln 2} \approx 1 - \frac{1.1}{0.69} = -\frac{41}{69} \approx -0.59 $

Which is quite a good approximation

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$\log_a(x) = y$

means by definition that

$a^y = x$

So to find $\log_2(0.667)$ you would (using the definition) need to solve the equation $2^y = 0.667$. We can solve that equation by "taking" $\log = \log_{10}$ on both sides, so we get

$\log(2^y) = \log(0.667) \Rightarrow$

$y\log(2) = \log(0.667) \Rightarrow$

$\log_2(0.667) = y = \frac{\log(0.667)}{\log(2)}$

To evaluate this expression you would need a calculator. Note you could also use the natural logarithm ($\ln = \log_{e}$, $e = 2.718281...$) and get

$\log_2(0.667) = \frac{\ln(0.667)}{\ln(2)}$

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