Recognize Logarithm Property: We are given log_(25)(1)/(125). The first step is to recognize that the logarithm of 1 to any base is always 0. log_b(1) = 0 for any base b. So, log_(25)(1) = 0.Express 1/125 as Power: Now, we need to express 1/125 as a power of 25. Since 125 is 5^3 and 25 is 5^2, we can write 1/125 as 5^(-3). However, we need it in terms of base 25, so we can write 5^(-3) as (5^2)^(-3/2) because (5^2)^(-3/2) = 5^(-3). Therefore, 1/125 can be written as 25^(-3/2).Rewrite Using New Expression: Now we can rewrite the logarithm using the new expression for 1/125. log_(25)(1)/(125) becomes log_(25)(25^(-3/2)).Simplify Using Logarithm Property: Using the property of logarithms that log_b(b^x) = x, we can simplify the expression. log_(25)(25^(-3/2)) = -3/2.

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$\log _{25} \frac{1}{125}=$

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Algebra 2

Evaluate logarithms using properties

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\log _{25} \frac{1}{125}=

Recognize Logarithm Property: We are given $\log_{25}\left(\frac{1}{125}\right)$ . The first step is to recognize that the logarithm of $1$ to any base is always $0$ . $\newline$ $\log_b(1) = 0$ for any base $b$ . $\newline$ So, $\log_{25}(1) = 0$ .
Express $\frac{1}{125}$ as Power: Now, we need to express $\frac{1}{125}$ as a power of $25$ . Since $125$ is $5^3$ and $25$ is $5^2$ , we can write $\frac{1}{125}$ as $5^{-3}$ . However, we need it in terms of base $25$ , so we can write $5^{-3}$ as $\frac{1}{125}$ $1$ because $\frac{1}{125}$ $2$ . Therefore, $\frac{1}{125}$ can be written as $\frac{1}{125}$ $4$ .
Rewrite Using New Expression: Now we can rewrite the logarithm using the new expression for $\frac{1}{125}$ . $\log_{25}\left(\frac{1}{125}\right)$ becomes $\log_{25}\left(25^{-\frac{3}{2}}\right)$ .
Simplify Using Logarithm Property: Using the property of logarithms that $\log_b(b^x) = x$ , we can simplify the expression. $\newline$ $\log_{25}(25^{-\frac{3}{2}}) = -\frac{3}{2}.$