What is the approximate value of $\log _6 25$?
We are tasked with finding the approximate value of the logarithm \(\log_6 25\). The expression \(\log_6 25\) represents the power to which the base \(6\) must be raised to obtain \(25\). Let \(x = \log_6 25\). This is equivalent to saying \(6^x = 25\). By observing the powers of \(6\), we know \(6^1 = 6\) and \(6^2 = 36\). Since \(6 < 25 < 36\), the exponent \(x\) must be a value between \(1\) and \(2\). Comparing this to our given options (a: \(0.233\), b: \(0.557\), c: \(0.620\), d: \(1.796\)), only one value falls within the interval \((1, 2)\).
To precisely calculate the value of \(\log_6 25\), we use the change-of-base formula, which states that \(\log_b x = \frac{\ln x}{\ln b}\) or \(\log_b x = \frac{\log_{10} x}{\log_{10} b}\). Applying this formula gives:
\[\log_6 25 = \frac{\ln 25}{\ln 6}\]
Using a calculator or logarithmic tables, we find \(\ln 25 \approx 3.21888\) and \(\ln 6 \approx 1.79176\). Dividing these two values yields:
\[\frac{3.21888}{1.79176} \approx 1.79649\]
Rounding this to three decimal places gives \(1.796\).
Comparing our result of approximately \(1.796\) with the given options, we find that it perfectly matches option (d). Therefore, the approximate value of \(\log_6 25\) is \(1.796\).
\[\log_6 25 \approx \boxed{1.796}\]
Logarithms are essential in measuring growth rates and determining time intervals in various fields. For example, in finance, logarithms are used to determine how long it takes for an investment to grow to a certain amount, similar to finding the exponent \(x\) in \(6^x = 25\). This is vital for calculating compound interest, financial planning, and understanding how quickly assets appreciate over time. In science, they are used to measure magnitudes of physical quantities like sound intensity (decibels) or the pH of a substance, where values cover a very broad range.