What would be the outcome of simplifying the expression (logs 9) × (√4) 10836 8, given that log2 3 = k?

Let's walk through the simplification of the given expression carefully. Here is the original expression:

\[
(\log_s 9) \times (\sqrt{4}) \times 10836 \times 8
\]

It seems there is some ambiguity regarding the notation “\(\log_s 9\).” I’ll assume that it should be interpreted as either \(\log_2(9)\) or \(\log(9)\) (common logarithm, base 10). Since the problem gives that \(\log_2 3 = k\), the most reasonable assumption is that it refers to a **logarithm in base 2**.

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### Step 1: Express \(\log_2(9)\) in terms of \(k\)
Since \(9 = 3^2\), we can apply logarithmic rules:

\[
\log_2(9) = \log_2(3^2) = 2 \cdot \log_2(3) = 2k
\]

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### Step 2: Simplify the square root term
The expression \(\sqrt{4}\) evaluates to:

\[
\sqrt{4} = 2
\]

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### Step 3: Rewrite the expression
Now, substitute the results into the original expression:

\[
(2k) \times 2 \times 10836 \times 8
\]

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### Step 4: Simplify the multiplication
First, compute the numerical products step by step:

\[
2 \times 8 = 16
\]

\[
16 \times 10836 = 173376
\]

Thus, the expression becomes:

\[
2k \times 173376
\]

Finally:

\[
346752k
\]

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### Final Answer:
The simplified expression is:

\[
346752k
\]

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Would you like a deeper breakdown of any specific step or more explanations?  

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**Related Questions:**
1. How do you simplify logarithmic expressions involving roots and powers?
2. What are the rules for converting between logarithmic bases?
3. How can \(\log_2(9)\) be expressed if we only know \(\log(3)\) (base 10)?
4. What would happen if \(\log_s 9\) were interpreted as a different base?
5. How do logarithmic identities help in solving complex equations?

**Tip:** When simplifying logarithmic terms, always look for ways to express them using known variables or constants.

Find out how to simplify the expression (log2 9) × (√4) × 10836 × 8 given log2 3 = k. This solution breaks down the logarithmic manipulation, properties of exponents, and algebraic steps. Suitable for high school students in Grades 9-12.

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