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Chapter 13: Problem 37

Write as the sum or difference of logarithms and simplify, if possible. Assume all variables represent positive real numbers. $$\log _{7} \frac{a^{2}}{b^{5}}$$

Short Answer

Expert verified
\( \log_{7} \frac{a^{2}}{b^{5}} = 2\log_{7}(a) - 5\log_{7}(b) \)

Step by step solution

01

Break the logarithm of a fraction into a difference of two logarithms

To break the logarithm of a fraction, we use the property of logarithms: \(\log_{a}(b/c) = \log_{a}(b) - \log_{a}(c)\). Therefore, we have \[ \log_{7} \frac{a^{2}}{b^{5}} = \log_{7}(a^2) - \log_{7}(b^5). \]
02

Simplify the logarithm using logarithmic properties

Now, we use the power property of logarithms: \(\log_{a}(b^c) = c \log_{a}(b)\). Apply this property to both logarithmic terms: \[ \log_{7}(a^2) - \log_{7}(b^5) = 2\log_{7}(a) - 5\log_{7}(b). \] The expression is now written as a sum or difference of logarithms, and there are no more possible simplifications. Hence, the final answer is: \[ \log_{7} \frac{a^{2}}{b^{5}} = 2\log_{7}(a) - 5\log_{7}(b). \]

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