Problem 109 - College Algebra

Question

Describe the change-of-base property and give an example.

Step-by-Step Solution

Answer

The change-of-base property for logarithms states that \(\log_b x = \frac{\log_a x}{\log_a b}\) for any positive numbers \(a\), \(b\), \((b≠1)\) and \(x\). This property helps to calculate logarithms with bases that are not easily computable. An example of this: \(\log_{2} 64 = \frac{\log_ {10} 64}{\log_ {10} 2} ≈ 6\).

Step 1: Understand the change-of-base property

The change-of-base property of logarithms states that for any positive number \(a\), \(b\), \((b≠1)\) and \(x\), the logarithm base \(b\) of \(x\) can be computed with the help of logarithms with another base \(a\). This is especially helpful when dealing with bases that are not on a calculator. It is generally written as: \(\log_b x = \frac{\log_a x}{\log_a b}\). Note that the base \(a\) in the new expression can be any positive value except 1.

Step 2: Provide the derivation of the formula

Suppose \(y = b^x\), then take the logarithm of both sides using the base \(a\), we will get \(\log_a y = x \cdot log_a b\). if \(y=x\), we will get \(\log_a x = x \cdot log_a b\) hence, \(x = \frac{\log_a x}{\log_a b}\). This is how we can arrive at the change-of-base formula.

Step 3: Example using change-of-base property

For example, consider calculating \(\log_{2} 64\). Using the change-of-base property, this can transform into \(\frac{\log_ {10} 64}{\log_ {10} 2}\), which are values we can easily calculate with a calculator. So, \(\log_{2} 64 = \frac{\log_ {10} 64}{\log_ {10} 2} = \frac{1.806}{0.301} ≈ 6\).