Question - Solving Logarithmic Expressions with Log Rules
Solution:
To solve the expression given in the image, $$ \log_{c^2}(\frac{a^5b}{c^2}) $$, we can apply log rules (quotient rule, power rule, and change of base formula).The quotient rule states that $$ \log(\frac{x}{y}) = \log(x) - \log(y) $$.The power rule states that $$ \log(x^a) = a \log(x) $$.The change of base formula states that $$ \log_b(a) = \frac{\log_c(a)}{\log_c(b)} $$.Let's break down the expression by using these rules:1. Apply the quotient rule to the argument of the log. $$ \log_{c^2}(a^5b) - \log_{c^2}(c^2) $$2. Apply the power rule to $$ a^5 $$ and $$ c^2 $$ within the log expressions. $$ 5 \log_{c^2}(a) + \log_{c^2}(b) - 2 \log_{c^2}(c) $$3. Notice that $$ \log_{c^2}(c^2) = 1 $$ because the log base and the argument are the same value raised to the same power. Hence, the expression becomes: $$ 5 \log_{c^2}(a) + \log_{c^2}(b) - 2 $$4. Apply the change of base formula to the two remaining log terms. Since we want to express everything in terms of the base c, we get: $$ 5 \frac{\log_{c}(a)}{\log_{c}(c^2)} + \frac{\log_{c}(b)}{\log_{c}(c^2)} - 2 $$5. Simplify by recognizing that $$ \log_{c}(c^2) = 2 $$. $$ 5 \frac{\log_{c}(a)}{2} + \frac{\log_{c}(b)}{2} - 2 $$6. Simplify the fractions. $$ \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 $$7. Looking at the answer choices given, we can see that our derived expression matches choice D when we distribute the 2 outside of the log: $$ \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \cdot 1 $$ $$ \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \log_{c}(c) $$Since $$ \log_{c}(c) = 1 $$, the 2 just becomes $$ 2 \log_{c}(c) $$, and thus the answer is: $$ \frac{5}{2} \cdot \log_{c}(a) + \frac{1}{2} \cdot \log_{c}(b) - 2 \log_{c}(c) $$So, the correct answer is D: $$ 5 \log_{c}(a) + \log_{c}(b) - 2 \log_{c}(c) $$
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