Write the expression as a single logarithm.  Express powers as factors.
log Subscript 3StartRoot x EndRootminuslog Subscript 3x Superscript 4
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Part 1
log Subscript 3StartRoot x EndRootminuslog Subscript 3x Superscript 4equals

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​(Type an exact answer. Use integers or fractions for any numbers in the​ expression.)

To simplify the expression \( \log_3 \sqrt{x} - \log_3 x^4 \) as a single logarithm, follow these steps:

1. **Rewrite the square root and exponent as powers:**
   - \( \sqrt{x} = x^{1/2} \), so \( \log_3 \sqrt{x} = \log_3 x^{1/2} \).
   - We can also rewrite \( x^4 \) in the second term directly.

   Now the expression becomes:
   \[
   \log_3 x^{1/2} - \log_3 x^4
   \]

2. **Apply the power rule of logarithms** \( \log_b a^c = c \cdot \log_b a \):
   - Rewrite each term as:
     \[
     \frac{1}{2} \log_3 x - 4 \log_3 x
     \]

3. **Combine terms:**
   - Now, we have:
     \[
     \frac{1}{2} \log_3 x - 4 \log_3 x = \left( \frac{1}{2} - 4 \right) \log_3 x
     \]

4. **Simplify the expression in parentheses:**
   - \(\frac{1}{2} - 4 = \frac{1}{2} - \frac{8}{2} = -\frac{7}{2}\)

   So the expression simplifies to:
   \[
   -\frac{7}{2} \log_3 x
   \]

### Final Answer:
\[
\log_3 \sqrt{x} - \log_3 x^4 = -\frac{7}{2} \log_3 x
\]

Would you like a further breakdown, or have any questions?

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**Related Practice Questions:**
1. Simplify \( \log_5 \sqrt{y} - \log_5 y^3 \).
2. Express \( \log_2 8 + \log_2 16 \) as a single logarithm.
3. Rewrite \( 2 \log_4 x - \log_4 (x + 3) \) as a single logarithm.
4. Simplify \( \log_7 (x^5) - \frac{1}{3} \log_7 y \).
5. Combine \( \log_9 a + 3 \log_9 b - \log_9 c \) into a single logarithm.

**Tip:** When combining logarithmic expressions, look for the opportunity to use the power, product, and quotient rules to simplify.

Write the expression as a single logarithm. Express powers as factors.

log_3(sqrt(x)) - log_3(x^4)

Learn how to simplify the expression log_3(sqrt(x)) - log_3(x^4) using logarithmic properties, including the power and difference of logarithms rules. This step-by-step solution is suitable for students in Grades 9-11.

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