log_(3)(-cos x)-log_(9)(sin x)+(1)/(4)=-log_(9)2

Convert to Base 3: Convert the base of the second logarithm from base 9 to base 3 using the change of base formula: log_b(a) = log_c(a) / log_c(b). Calculation: log_(9)(sin x) = log_(3)(sin x) / log_(3)(9) Since log_(3)(9) = 2, it simplifies to log_(9)(sin x) = (1/2) * log_(3)(sin x).Substitute Converted Logarithm: Substitute the converted logarithm back into the original equation. Calculation: log_(3)(-cos x) - (1/2) * log_(3)(sin x) + 1/4 = -log_(9)(2)Convert to Base 3: Convert -log_(9)(2) to base 3 using the same change of base formula. Calculation: -log_(9)(2) = -log_(3)(2) / log_(3)(9) Since log_(3)(9) = 2, it simplifies to -log_(9)(2) = -(1/2) * log_(3)(2).Combine Terms in Base 3: Combine all terms in base 3. Calculation: log_(3)(-cos x) - (1/2) * log_(3)(sin x) + 1/4 = -(1/2) * log_(3)(2)Isolate Logarithm: Attempt to isolate log_(3)(-cos x). Calculation: log_(3)(-cos x) = (1/2) * log_(3)(sin x) - 1/4 + (1/2) * log_(3)(2)

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log_(3)(-cos x)-log_(9)(sin x)+(1)/(4)=-log_(9)2

$\log _{3}(-\cos x)-\log _{9}(\sin x)+\frac{1}{4}=-\log _{9} 2$

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Precalculus

Quotient property of logarithms

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\log _{3}(-\cos x)-\log _{9}(\sin x)+\frac{1}{4}=-\log _{9} 2

Convert to Base $3$ : Convert the base of the second logarithm from base $9$ to base $3$ using the change of base formula: $\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$ . $\newline$ Calculation: $\log_{9}(\sin x) = \frac{\log_{3}(\sin x)}{\log_{3}(9)}$ $\newline$ Since $\log_{3}(9) = 2$ , it simplifies to $\log_{9}(\sin x) = \frac{1}{2} * \log_{3}(\sin x)$ .
Substitute Converted Logarithm: Substitute the converted logarithm back into the original equation. $\newline$ Calculation: $\log_{3}(-\cos x) - \frac{1}{2} \cdot \log_{3}(\sin x) + \frac{1}{4} = -\log_{9}(2)$
Convert to Base $3$ : Convert $-\log_{9}(2)$ to base $3$ using the same change of base formula. $\newline$ Calculation: $-\log_{9}(2) = -\frac{\log_{3}(2)}{\log_{3}(9)}$ $\newline$ Since $\log_{3}(9) = 2$ , it simplifies to $-\log_{9}(2) = -\left(\frac{1}{2}\right) \cdot \log_{3}(2)$ .
Combine Terms in Base $3$ : Combine all terms in base $3$ . $\newline$ Calculation: $\log_{3}(-\cos x) - \frac{1}{2} \cdot \log_{3}(\sin x) + \frac{1}{4} = -\frac{1}{2} \cdot \log_{3}(2)$
Isolate Logarithm: Attempt to isolate $\log_{3}(-\cos x)$ . $\newline$ Calculation: $\log_{3}(-\cos x) = \frac{1}{2} \cdot \log_{3}(\sin x) - \frac{1}{4} + \frac{1}{2} \cdot \log_{3}(2)$