To prove that \( \log_{\sqrt{b}} x = 2 \log_b x \), we'll use properties of logarithms and exponents.

### Proof:

1. **Start with the expression:**
   \[
   \log_{\sqrt{b}} x
   \]

2. **Express the logarithm with the square root base as an exponent:**
   \[
   \log_{\sqrt{b}} x = \log_{b^{1/2}} x
   \]
   Since \( \sqrt{b} \) is the same as \( b^{1/2} \).

3. **Use the change of base property of logarithms:**
   \[
   \log_{b^a} x = \frac{1}{a} \log_b x
   \]
   In this case, \( a = \frac{1}{2} \), so we can rewrite:
   \[
   \log_{b^{1/2}} x = \frac{1}{\frac{1}{2}} \log_b x = 2 \log_b x
   \]

Thus, we have proven that:
\[
\log_{\sqrt{b}} x = 2 \log_b x
\]

### Conclusion:
The equation \( \log_{\sqrt{b}} x = 2 \log_b x \) is valid, as shown using the properties of logarithms.

---

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How can the change of base formula be applied to simplify complex logarithms?
2. What is the logarithmic relationship when changing the base from \( b \) to \( b^k \)?
3. How do logarithmic properties handle fractional and negative bases?
4. Can you solve for \( x \) in an equation involving \( \log_b x \) if \( b \) is a square root?
5. How does the logarithm of a product or quotient relate to logarithms of individual terms?

**Tip:** Always remember that the logarithm of a number to an exponent can be simplified using the power rule: \( \log_b (x^n) = n \log_b x \).

Explore the proof of the logarithmic identity log base √b(x) = 2 log base b(x) using the change of base formula and exponent rules. Learn how logarithmic properties simplify complex expressions. Suitable for Grades 10-12.

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