Problem 70 - Introductory and Intermediate Algebra for College Students 4th

Question

Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\) Write each expression in terms of \(A\) and \(C\). $$\log _{b} 6$$

Step-by-Step Solution

Answer

\(\log_b 6 = A + C\)

Step 1: Express 6 as a product of 2 and 3

Firstly, we can express 6 as a product of 2 and 3, i.e., \(6 = 2 \cdot 3\). This is done because we have \(\log_b 2 = A\) and \(\log_b 3 = C\) in terms of logarithms and our target is to express \(\log_b 6\) in terms of A and C.

Step 2: Application of Logarithm Product Rule

Next, we will apply the product rule of logarithms to \(\log_b 6\). This rule states that \(\log_b(m \cdot n) = \log_b m + \log_b n\). In this case, \(m = 2 \) and \(n = 3\). So, \(\log_b 6 = \log_b (2 \cdot 3) = \log_b 2 + \log_b 3\).

Step 3: Replacing Logs with their equivalents

Since \(\log_b 2 = A\) and \(\log_b 3 = C\), we can replace \(\log_b 2\) and \(\log_b 3\) with their equivalents A and C respectively in the above expression. This implies that \(\log_b 6 = A + C\).