Solve $\log _2 \frac{1}{64}$.
To solve the expression \(\log _2 \frac{1}{64}\), we first define the logarithm. The expression \(\log_b x = y\) is equivalent to the exponential equation \(b^y = x\). In this problem, the base \(b\) is \(2\) and the argument \(x\) is \(\frac{1}{64}\). Therefore, we want to find the value of \(y\) such that \(2^y = \frac{1}{64}\).
Next, we express \(\frac{1}{64}\) as a power of \(2\). We know that \(64 = 2^6\). Using the property of negative exponents, which states that \(\frac{1}{a^n} = a^{-n}\), we can rewrite the fraction:
\[\frac{1}{64} = \frac{1}{2^6} = 2^{-6}.\]
Finally, we substitute this back into our exponential equation:
\[2^y = 2^{-6}.\]
Since the bases are equal, the exponents must also be equal. Therefore, \(y = -6\). Thus, \(\log_2 \frac{1}{64} = -6\).
Logarithms are used in various fields, such as measuring the intensity of sounds in decibels or the magnitude of earthquakes using the Richter scale. For instance, if an earthquake's intensity is 1,000,000 times that of a reference earthquake, the magnitude is given by \(\log_{10}(1,000,000) = 6\). Understanding logarithmic properties like \(\log_b (\frac{1}{x^n}) = -n \log_b x\) helps in scaling data efficiently and solving problems where values grow or decay exponentially.