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.

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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: $\log_b (x^n) = n \log_b x$.