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$\log \frac{14}{3}+\log \frac{11}{5}-\log \frac{22}{15}=\log$ [ ]
We are given the logarithmic equation \(\log \frac{14}{3} + \log \frac{11}{5} - \log \frac{22}{15} = \log X\). Our objective is to determine the value of \(X\) by using the fundamental properties of logarithms. These properties include the product rule \(\log A + \log B = \log(A \cdot B)\) and the quotient rule \(\log A - \log B = \log(\frac{A}{B})\).
First, we apply the product rule to the first two terms of the expression:
\(\log \frac{14}{3} + \log \frac{11}{5} = \log \left( \frac{14}{3} \cdot \frac{11}{5} \right)\).
Multiplying the fractions inside the logarithm:
\(\frac{14}{3} \cdot \frac{11}{5} = \frac{14 \times 11}{3 \times 5} = \frac{154}{15}\).
So, the equation becomes \(\log \frac{154}{15} - \log \frac{22}{15} = \log X\).
Next, we apply the quotient rule to subtract the remaining term:
\(\log \frac{154}{15} - \log \frac{22}{15} = \log \left( \frac{\frac{154}{15}}{\frac{22}{15}} \right)\).
Simplifying the complex fraction inside the logarithm:
\(\frac{\frac{154}{15}}{\frac{22}{15}} = \frac{154}{15} \cdot \frac{15}{22} = \frac{154}{22} = 7\).
Therefore, the equation simplifies to \(\log 7 = \log X\).
Since \(\log 7 = \log X\), it follows that the argument of the logarithm must be equal, so \(X = 7\). Thus, the missing value is \(\boxed{7}\).
Logarithms are frequently used in fields like acoustics to measure sound intensity levels (in decibels) or in chemistry to calculate the pH level of a substance, which is defined as the negative base-10 logarithm of the concentration of hydronium ions. Understanding how to combine or split logarithmic expressions allows scientists and engineers to simplify complex formulas, making them easier to compute and analyze, such as when comparing the relative strengths of two different sound sources or acidity levels.