Question

log0.5 (1/8)

Answer 1

To calculate \(\log_{0.5} \left(\frac{1}{8}\right)\), we can use the change of base formula or properties of logarithms.

First, note that:

\[
\frac{1}{8} = 8^{-1} = (2^3)^{-1} = 2^{-3}
\]

Now we can express \(\log_{0.5} \left(\frac{1}{8}\right)\) in terms of base 2.

Since \(0.5 = 2^{-1}\), we can use the change of base formula. We know:

\[
\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}
\]

Using base \(2\):

\[
\log_{0.5}\left(\frac{1}{8}\right) = \frac{\log_{2}\left(\frac{1}{8}\right)}{\log_{2}(0.5)}
\]

We already established that \(\frac{1}{8} = 2^{-3}\), so:

\[
\log_{2} \left(\frac{1}{8}\right) = \log_{2} (2^{-3}) = -3
\]

Also, since \(0.5 = 2^{-1}\):

\[
\log_{2}(0.5) = \log_{2}(2^{-1}) = -1
\]

Now substituting these values back in:

\[
\log_{0.5}\left(\frac{1}{8}\right) = \frac{-3}{-1} = 3
\]

Thus, the value of \(\log_{0.5}\left(\frac{1}{8}\right)\) is \(3\).