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log_(81)(1)/(27)=

log81127= \log _{81} \frac{1}{27}=

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Q. log81127= \log _{81} \frac{1}{27}=
  1. Recognize base and argument: Recognize the base of the logarithm and the argument. The base of the logarithm is 8181, and the argument is 127\frac{1}{27}. We need to express 127\frac{1}{27} as a power of 8181.
  2. Express as power of 8181: Express 127\frac{1}{27} as a power of 8181.\newlineSince 8181 is 343^4 and 2727 is 333^3, we can write 127\frac{1}{27} as (34)13(3^4)^{-\frac{1}{3}} because (34)13=343=27(3^4)^{\frac{1}{3}} = 3^{\frac{4}{3}} = 27.
  3. Rewrite using new expression: Rewrite the logarithm using the new expression for 127\frac{1}{27}. \newlinelog81((34)13)\log_{81}\left((3^4)^{-\frac{1}{3}}\right)
  4. Apply change of base formula: Apply the change of base formula for logarithms.\newlineThe change of base formula is logb(a)=logc(a)logc(b)\log_b(a) = \frac{\log_c(a)}{\log_c(b)}. In this case, we can use base 33 for both the numerator and the denominator.\newlinelog81((34)13)=log3((34)13)log3(81)\log_{81}((3^4)^{-\frac{1}{3}}) = \frac{\log_3((3^4)^{-\frac{1}{3}})}{\log_3(81)}
  5. Simplify logarithms: Simplify the logarithms.\newlineSince 8181 is 343^4, log3(81)=4\log_3(81) = 4. The numerator is log3((34)1/3)\log_3((3^4)^{-1/3}), which simplifies to 13×log3(34)-\frac{1}{3} \times \log_3(3^4) because of the power property of logarithms.
  6. Simplify numerator further: Simplify the numerator further.\newline13×log3(34)-\frac{1}{3} \times \log_3(3^4) simplifies to 13×4-\frac{1}{3} \times 4 because log3(34)=4\log_3(3^4) = 4.
  7. Calculate expression value: Calculate the value of the expression. \newline13×4/4-\frac{1}{3} \times 4 / 4 simplifies to 13-\frac{1}{3}.
  8. Conclude final answer: Conclude the final answer.\newlineThe value of log81(127)\log_{81}(\frac{1}{27}) is 13-\frac{1}{3}.

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