Evaluate: log_(32)((1)/(64)) Answer:

Understand and Identify Properties: Understand the problem and identify the properties of logarithms to use. We need to evaluate the logarithm of 1/64 with base 32. We can use the change of base formula for logarithms, which is log base b of a equals log a divided by log b, where all the logarithms are in the same base.Apply Change of Base Formula: Apply the change of base formula. Using the change of base formula, we can write the expression as: log_(32)((1)/(64)) = log((1)/(64)) / log(32) We will use the common logarithm (base 10) for this calculation.Evaluate Using Known Values: Evaluate the logarithms using known values. We know that 32 is 2^5 and 64 is 2^6. Therefore, we can rewrite the logarithms as: log((1)/(64)) = log(1) - log(64) = 0 - log(2^6) = -6 * log(2) log(32) = log(2^5) = 5 * log(2)Divide Logarithms: Divide the two logarithms. Now we divide the two expressions we found in Step 3: (-6 * log(2)) / (5 * log(2)) Since log(2) is in both the numerator and the denominator, they cancel out, leaving us with: -6 / 5

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Evaluate: $\newline$ $\log _{32} \frac{1}{64}$ $\newline$ Answer:

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Precalculus

Quotient property of logarithms

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Q. Evaluate:

\newline

\log _{32} \frac{1}{64}

\newline

Answer:

Understand and Identify Properties: Understand the problem and identify the properties of logarithms to use. $\newline$ We need to evaluate the logarithm of $\frac{1}{64}$ with base $32$ . We can use the change of base formula for logarithms, which is $\log_{b}a = \frac{\log a}{\log b}$ , where all the logarithms are in the same base.
Apply Change of Base Formula: Apply the change of base formula. $\newline$ Using the change of base formula, we can write the expression as: $\newline$ $\log_{32}\left(\frac{1}{64}\right) = \frac{\log\left(\frac{1}{64}\right)}{\log(32)}$ $\newline$ We will use the common logarithm (base $10$ ) for this calculation.
Evaluate Using Known Values: Evaluate the logarithms using known values. $\newline$ We know that $32$ is $2^5$ and $64$ is $2^6$ . Therefore, we can rewrite the logarithms as: $\newline$ $log\left(\frac{1}{64}\right) = log(1) - log(64) = 0 - log(2^6) = -6 \cdot log(2)$ $\newline$ $log(32) = log(2^5) = 5 \cdot log(2)$
Divide Logarithms: Divide the two logarithms. $\newline$ Now we divide the two expressions we found in Step $3$ : $\newline$ $(-6 \times \log(2)) / (5 \times \log(2))$ $\newline$ Since $\log(2)$ is in both the numerator and the denominator, they cancel out, leaving us with: $\newline$ $-6 / 5$