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

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

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Evaluate logarithms using properties

Full solution

Q.

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

Recognize base and argument: Recognize the base of the logarithm and the argument. The base of the logarithm is $81$ , and the argument is $\frac{1}{27}$ . We need to express $\frac{1}{27}$ as a power of $81$ .
Express as power of $81$ : Express $\frac{1}{27}$ as a power of $81$ . $\newline$ Since $81$ is $3^4$ and $27$ is $3^3$ , we can write $\frac{1}{27}$ as $(3^4)^{-\frac{1}{3}}$ because $(3^4)^{\frac{1}{3}} = 3^{\frac{4}{3}} = 27$ .
Rewrite using new expression: Rewrite the logarithm using the new expression for $\frac{1}{27}$ . $\newline$ $\log_{81}\left((3^4)^{-\frac{1}{3}}\right)$
Apply change of base formula: Apply the change of base formula for logarithms. $\newline$ The change of base formula is $\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$ . In this case, we can use base $3$ for both the numerator and the denominator. $\newline$ $\log_{81}((3^4)^{-\frac{1}{3}}) = \frac{\log_3((3^4)^{-\frac{1}{3}})}{\log_3(81)}$
Simplify logarithms: Simplify the logarithms. $\newline$ Since $81$ is $3^4$ , $\log_3(81) = 4$ . The numerator is $\log_3((3^4)^{-1/3})$ , which simplifies to $-\frac{1}{3} \times \log_3(3^4)$ because of the power property of logarithms.
Simplify numerator further: Simplify the numerator further. $\newline$ $-\frac{1}{3} \times \log_3(3^4)$ simplifies to $-\frac{1}{3} \times 4$ because $\log_3(3^4) = 4$ .
Calculate expression value: Calculate the value of the expression. $\newline$ $-\frac{1}{3} \times 4 / 4$ simplifies to $-\frac{1}{3}$ .
Conclude final answer: Conclude the final answer. $\newline$ The value of $\log_{81}(\frac{1}{27})$ is $-\frac{1}{3}$ .

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