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Write as the sum and/or difference of logarithms. Express powers as factors. \( \log _{8}\left(\frac{x y^{9}}{z^{7}}\right) \)

Ask by Huff Bolton. in the United States
Mar 15,2025

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\[ \log_{8}(x) + 9\log_{8}(y) - 7\log_{8}(z) \]

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First, we can use the logarithm of a quotient rule, which states that \( \log_b\left(\frac{A}{B}\right) = \log_b(A) - \log_b(B) \). Here, we can apply this to rewrite the logarithm of the fraction. Next, we can apply the logarithm of a product rule, which states that \( \log_b(A \cdot C) = \log_b(A) + \log_b(C) \), and the power rule, which states that \( \log_b(A^n) = n \cdot \log_b(A) \). So, we start with: \[ \log_{8}\left(\frac{x y^{9}}{z^{7}}\right) = \log_{8}(x y^{9}) - \log_{8}(z^{7}) \] Now applying the product rule: \[ \log_{8}(x y^{9}) = \log_{8}(x) + \log_{8}(y^{9}) \] Now applying the power rule: \[ \log_{8}(y^{9}) = 9 \cdot \log_{8}(y) \] Putting it all together: \[ \log_{8}\left(\frac{x y^{9}}{z^{7}}\right) = \log_{8}(x) + 9 \cdot \log_{8}(y) - 7 \cdot \log_{8}(z) \] Thus, the final expression is: \[ \log_{8}(x) + 9 \cdot \log_{8}(y) - 7 \cdot \log_{8}(z) \]

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