### Instruction: Find the positive value of $x$ which satisfies
\[\log_5 (x - 2) + \log_{\sqrt{5}} (x^3 - 2) + \log_{\frac{1}{5}} (x - 2) = 4.\].### Response: This problem involves three logarithms with different bases, so I want to find a way to simplify them or use some properties of logarithms to rewrite them.
I notice that the bases of the logarithms are related by powers of 5: $\sqrt{5} = 5^{1/2}$ and $\frac{1}{5} = 5^{-1}$, so I can use the change of base formula to write them all in terms of $\log_5$.
The change of base formula says that $\log_b a = \frac{\log_c a}{\log_c b}$, where $c$ is any positive base.
In this case, I can choose $c = 5$ and write
\[\log_{\sqrt{5}} (x^3 - 2) = \frac{\log_5 (x^3 - 2)}{\log_5 \sqrt{5}} = \frac{\log_5 (x^3 - 2)}{\frac{1}{2}} = 2 \log_5 (x^3 - 2)\]
and
\[\log_{\frac{1}{5}} (x - 2) = \frac{\log_5 (x - 2)}{\log_5 \frac{1}{5}} = \frac{\log_5 (x - 2)}{-1} = -\log_5 (x - 2).\]
Substituting these expressions into the original equation, I get
\[\log_5 (x - 2) + 2 \log_5 (x^3 - 2) - \log_5 (x - 2) = 4.\]
Simplifying the left-hand side, I see that the first and last terms cancel out, leaving me with
\[2 \log_5 (x^3 - 2) = 4.\]
To isolate the logarithm, I divide both sides by 2 and get
\[\log_5 (x^3 - 2) = 2.\]
To get rid of the logarithm, I exponentiate both sides with base 5 and get
\[x^3 - 2 = 5^2.\]
Expanding the right-hand side, I get
\[x^3 - 2 = 25.\]
To solve for $x$, I add 2 to both sides and take the cube root and get
\[x = \sqrt[3]{27}.\]
This is the positive value of $x$ that satisfies the equation.
I could check by plugging it back into the original equation and verifying that both sides are equal.