evaluate each expression.

$\log_{3}27 = $

$\log_{12}1 = $

$\log_{5} \frac{1}{25} = $

$\log_{2}128 = $

Question

evaluate each expression.

$\log_{3}27 = $

$\log_{12}1 = $

$\log_{5} \frac{1}{25} = $

$\log_{2}128 = $

Accepted Answer

{ "quality": "clear", "question_count": 4, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Logarithmic Functions" ], "new_concepts": [ "Logarithmic Evaluation" ], "current_concepts": [ "Logarithmic Functions", "Logarithmic Evaluation" ] } ## Evaluate the first logarithm Using the *Logarithmic Functions* knowledge point $$ \begin{aligned} &\log_{3}27 = x \implies 3^x = 27\ &3^x = 3^3 \implies x = 3 \end{aligned} $$ ## Evaluate the second logarithm Using the *Logarithmic Functions* knowledge point $$ \begin{aligned} &\log_{12}1 = y \implies 12^y = 1\ &12^y = 12^0 \implies y = 0 \end{aligned} $$ ## Evaluate the third logarithm Using the *Logarithmic Functions* knowledge point $$ \begin{aligned} &\log_{5}\frac{1}{25} = z \implies 5^z = \frac{1}{25}\ &5^z = 5^{-2} \implies z = -2 \end{aligned} $$ ## Evaluate the fourth logarithm Using the *Logarithmic Functions* knowledge point $$ \begin{aligned} &\log_{2}128 = w \implies 2^w = 128\ &2^w = 2^7 \implies w = 7 \end{aligned} $$ | No. | Problem | Answer | |---|---|---| | 1 | $\log_{3}27$ | $3$ | | 2 | $\log_{12}1$ | $0$ | | 3 | $\log_{5}\frac{1}{25}$ | $-2$ | | 4 | $\log_{2}128$ | $7$ | { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Algebra", "Logarithmic Evaluation" ] }

No.	Problem	Answer
2	\(\log_{12}1\)	\(0\)
3	\(\log_{5}\frac{1}{25}\)	\(-2\)
4	\(\log_{2}128\)	\(7\)

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QUESTION IMAGE

Question

Explanation:

Evaluate the first logarithm

Evaluate the second logarithm

Evaluate the third logarithm

Evaluate the fourth logarithm

Answer: