Use the Laws of Logarithms to evaluate the expression.
log(base3)(9^300)

We can use the laws of logarithms to simplify and evaluate the expression \( \log_3(9^{300}) \).

### Step-by-Step Solution:

1. **Rewrite the logarithmic expression:**

   \[
   \log_3(9^{300})
   \]

2. **Use the power rule of logarithms**, which states that \( \log_b(a^n) = n \cdot \log_b(a) \). Apply this rule to move the exponent 300 in front:

   \[
   300 \cdot \log_3(9)
   \]

3. **Simplify \( \log_3(9) \):**

   Since \( 9 = 3^2 \), we can rewrite \( \log_3(9) \) as \( \log_3(3^2) \). Using the power rule again:

   \[
   \log_3(3^2) = 2 \cdot \log_3(3)
   \]

4. **Evaluate \( \log_3(3) \):**

   By definition, \( \log_3(3) = 1 \), since 3 raised to the power of 1 is 3.

   \[
   \log_3(3^2) = 2 \cdot 1 = 2
   \]

5. **Substitute back:**

   Now we substitute \( \log_3(9) = 2 \) into the original expression:

   \[
   300 \cdot 2 = 600
   \]

### Final Answer:

\[
\log_3(9^{300}) = 600
\]

Would you like more details, or do you have any questions?

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Here are 5 related questions:

1. How would you evaluate \( \log_3(27^{100}) \)?
2. Can you simplify \( \log_5(25^{50}) \)?
3. What is \( \log_2(16^{20}) \) using logarithmic laws?
4. How can you use the change of base formula to evaluate \( \log_3(81) \)?
5. Can you express \( \log_3(x^5) \) in simpler form?

**Tip:** The power rule of logarithms is helpful when working with expressions that involve exponents, as it allows you to bring the exponent outside the logarithm.

Use the Laws of Logarithms to evaluate the expression. log(base3)(9^300)

Learn how to use the power rule of logarithms to evaluate the expression log(base3)(9^300). This detailed step-by-step solution simplifies the problem and covers logarithmic concepts and exponentiation, suitable for high school students.

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