/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Let \(\log _{b} 2=A\) and \(\log... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
\(\log_b 6 = A + C\)

Step by step solution

01

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.
02

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, \(\log_b 6 = \log_b (2 \cdot 3) = \log_b 2 + \log_b 3\).
03

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 \(\log_b 6 = A + C\).

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