Verified Answer :

  • Express the argument 125 as a power of the base: \(125 = 5^3\).
  • Apply the logarithm property \(\log_b(b^x) = x\) to simplify \(\log_5(5^3)\) to \(3\).
  • Multiply the result by the coefficient \(\frac{1}{3}\) to find the final value: \(\frac{1}{3} \times 3 = 1\).
  • The final result is \(\boxed{1}\).

Explanation

  1. Problem Analysis

To evaluate the expression \(\frac{1}{3} \log_5(125)\), we first identify the components. We have a logarithm with base 5 and an argument of 125, multiplied by a coefficient of \(\frac{1}{3}\). Our goal is to simplify this expression into a single numerical value.

  1. Expressing the Argument as a Power of the Base

We recognize that 125 is a power of 5. Since \(5^1 = 5\), \(5^2 = 25\), and \(5^3 = 5 \times 5 \times 5 = 125\), we can write 125 as \(5^3\). Substituting this into the logarithmic part of the expression gives us \(\log_5(5^3)\).

  1. Simplifying the Logarithm

We use the logarithmic property \(\log_b(b^x) = x\), which states that the logarithm of a base raised to a power is simply the exponent. Applying this to our expression, we have \(\log_5(5^3) = 3\). Thus, the entire expression simplifies to \(\frac{1}{3} \times 3\).

  1. Final Calculation and Result

Finally, we perform the multiplication: \(\frac{1}{3} \times 3 = 1\). Therefore, the value of the given expression is 1.

Examples

Logarithms are fundamental in measuring quantities that span many orders of magnitude. For instance, the Richter scale used to measure earthquake intensity is logarithmic. If an earthquake A is 100 times stronger than earthquake B, the difference in their magnitude on the Richter scale is simply \(\log_{10}(100) = 2\). Understanding logarithmic properties helps in comparing these intensities easily and calculating energy levels in various scientific fields.

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