What is the value of $\log _{625} 5$ ?
We need to evaluate the expression \(\log_{625} 5\). The definition of a logarithm states that \(\log_{b} x = y\) is equivalent to \(b^y = x\), provided that \(b > 0, b eq 1\), and \(x > 0\). In this specific problem, \(b = 625\) and \(x = 5\). Therefore, we are looking for the exponent \(y\) such that \(625^y = 5\).
To solve the equation \(625^y = 5\), we express both sides with the same base. We know that \(625\) can be written as a power of \(5\):
\[625 = 5 \times 5 \times 5 \times 5 = 5^4.\]
Substituting this into our equation gives:
\[(5^4)^y = 5^1.\]
Using the power rule for exponents, which states that \((a^m)^n = a^{m \times n}\), we can simplify the left-hand side of the equation:
\[5^{4y} = 5^1.\]
Since the bases are equal and both are greater than 0 and not equal to 1, their exponents must be equal:
\[4y = 1.\]
Dividing both sides of the equation by 4, we find:
\[y = \frac{1}{4}.\]
Thus, \(\log_{625} 5 = \frac{1}{4}\). Comparing this result with the given options (a. -4, b. 1/4, c. -1/4, d. 4), we see that the correct choice is b.
Logarithms are essential in measuring scales that involve vast ranges of values, such as the Richter scale for earthquakes or the pH scale for acidity. For instance, if you are analyzing a sound level change and encounter a logarithmic ratio involving power or intensity, you can use these properties to simplify the calculations. This allows scientists and engineers to convert complex multiplicative growth relationships into simple, additive ones, making it easier to interpret data across many orders of magnitude.