Problem 13 - Precalculus : Building Concepts and Connections

Question

In Exercises \(7-14,\) use \(\log 2 \approx 0.3010, \log 5 \approx 0.6990,\) and \(\log 7 \approx 0.8451\) to evaluate each logarithm without using a calculator. Then check your answer using a calculator. $$\log 125$$

Step-by-Step Solution

Answer

Based on the mentioned steps, the value of \(\log 125\) evaluated without a calculator is approximately \(2.0970\).

Step 1: Identify Factors of 125

The number 125 can be factored into \(5 \cdot 5 \cdot 5\) or \(5^3\). Therefore, \(\log 125\) can be rewritten as \(\log (5^3)\).

Step 2: Apply Logarithmic Properties

Using the logarithmic property , \(\log (a^n) = n \cdot \log a\), our logarithm can be further simplified to be \(3 \cdot \log 5\).

Step 3: Substitute Given Value

We now plug in the given value \(\log 5 = 0.6990\) into the equation. Therefore, \(3 \cdot \log 5\)= \(3 \cdot 0.6990 = 2.0970\).

Key Concepts

Logarithmic Properties

To understand how logarithms work and how to evaluate them, it's essential to familiarize yourself with their properties. Logarithmic properties are rules that govern the operations of logarithms, which are the inverse of exponentiation. For instance, one key property is the Product Rule, which states that the logarithm of a product is the sum of the logarithms of the factors:

\[\begin{equation}\log(ab) = \log(a) + \log(b)\end{equation}\]
Another important property is the Quotient Rule, which informs us that the logarithm of a quotient is the difference between the logarithm of the numerator and the denominator:

\[\begin{equation}\log\left(\frac{a}{b}\right) = \log(a) - \log(b)\end{equation}\]
Lastly, the Power Rule, as used in the example problem, expresses that the logarithm of a number raised to an exponent is equal to the exponent multiplied by the logarithm of the number:

\[\begin{equation}\log(a^n) = n \cdot \log(a)\end{equation}\]
Knowing these properties is crucial as they allow you to manipulate and simplify logarithms, making them easier to evaluate without a calculator.

Logarithm Without Calculator

Evaluating logarithms without a calculator can seem daunting, yet it's quite manageable with the help of logarithmic properties and some clever arithmetic. Firstly, recognize that certain log values, such as those for 2, 5, and 7, can be memorized or approximated for quick calculations. When faced with an expression like \(\log 125\), breaking it down into known components is the key.

This is achieved by factoring the number inside the logarithm to base numbers with known log values. As demonstrated in the problem, 125 is factored into \(5^3\), and the Power Rule of logarithms is applied. By turning the exponent into a multiplier, the evaluation becomes a straightforward multiplication: \(3 \cdot \log 5\). With the given value for \(\log 5\), the problem simplifies into a basic arithmetic operation, thus eliminating the need for a calculator. This method greatly improves accuracy when working by hand and deepens the student's conceptual understanding of logarithms.

Logarithm Simplification

Simplifying logarithms is a process that makes them more manageable and easier to evaluate. Begin by looking for ways to express the number inside the logarithm as a product or power of numbers with known logarithms. It's like decomposing the number into building blocks that are more familiar. As seen in the exercise, \(\log 125\) was simplified by recognizing that 125 equals \(5^3\).

After expressing the number in a simpler form, apply the logarithmic properties to break down the expression. For instance, here we use the Power Rule. After simplifying the logarithm to \(3 \cdot \log 5\), we substitute the known value of \(\log 5\). The simplification process reduces complex logarithmic calculations to basic multiplication or addition, which are much simpler to carry out without a calculator. This promotes a deeper comprehension of the logarithm concept and ensures that even without technology, students can confidently tackle logarithmic problems.