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Logarithmic functions are incredibly useful in mathematics for solving equations involving exponential growth and decay. These functions form the inverse of exponential functions. Essentially, if you have an exponential function like \( b^x = y \), the corresponding logarithmic function is \( \log_b y = x \). This powerful relationship helps us explore and solve problems across different scientific fields and mathematical interests. In the exercise, we are dealing with several logarithms where the base of the logarithm, \( b \), is greater than 1. The expressions like \( \log_b 2 \), \( \log_b 3 \), and \( \log_b 5 \) show us how many times we need to multiply the base \( b \) to reach 2, 3, or 5 respectively. This forms the backbone of solving the expressions you are given.

Understanding the properties of logarithms is crucial to simplifying expressions and solving equations like those in our exercise. These properties allow us to rewrite and manipulate logarithmic expressions in simpler terms, which makes them easier to solve.

• The product rule states \( \log_b (mn) = \log_b m + \log_b n \).
• The quotient rule shows \( \log_b (\frac{m}{n}) = \log_b m - \log_b n \).
• The power rule gives \( \log_b (m^n) = n \cdot \log_b m \).

Also important is the identity \( \log_a b \times \log_b a = 1 \), because of their inverse relationship. In our exercise, this identity is directly used to simplify expressions to 1, since the components are inverse functions of each other.

The change of base formula is a handy tool when you need to evaluate logarithms with bases that aren't directly programmable into a calculator, or when doing manual calculations. This formula states that:\[\log_b a = \frac{\log_c a}{\log_c b}\]This formula lets you convert a logarithm from one base to another base \( c \) of your choosing. Now given the setup of your exercise, this formula can transform complex logarithmic expressions into forms involving the bases we know, such as \( A, B, \) and \( C \). For example, if you needed \( \log_6 b \), and we know \( \log_b 6 \), we can use this property, along with previously discussed identities, to find a simpler expression. If you remember from our steps, once we calculated \( \log_b 6 \) as \( A + B \), we could find \( \log_6 b \) as \( \frac{1}{A + B} \) to ensure the product of these two expressions equals 1. Such interconnections are what make logarithms both fascinating and functionally significant.