### 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 is a logarithmic equation, so I need to somehow combine or eliminate the logarithms on the left side to isolate $x$.
I notice that the bases of the logarithms are all related to $5$.
In fact, $\sqrt{5} = 5^{\frac{1}{2}}$ and $\frac{1}{5} = 5^{-1}$.
This means I can use the change of base formula to rewrite the logarithms in terms of $\log_5$.
For example, \[\log_{\sqrt{5}} (x^3 - 2) = \frac{\log_5 (x^3 - 2)}{\log_5 (\sqrt{5})} = \frac{\log_5 (x^3 - 2)}{\frac{1}{2}} = 2\log_5 (x^3 - 2).\]
Similarly, \[\log_{\frac{1}{5}} (x - 2) = \frac{\log_5 (x - 2)}{\log_5 (\frac{1}{5})} = \frac{\log_5 (x - 2)}{-1} = -\log_5 (x - 2).\]
Substituting these expressions into the equation, I get
\[\log_5 (x - 2) + 2\log_5 (x^3 - 2) - \log_5 (x - 2) = 4.\]
Now I can use the properties of logarithms to simplify the left side.
For example, $\log_5 (x - 2) - \log_5 (x - 2) = 0$, and $2\log_5 (x^3 - 2) = \log_5 ((x^3 - 2)^2)$.
This gives me
\[\log_5 ((x^3 - 2)^2) = 4.\]
To get rid of the logarithm, I can exponentiate both sides with base $5$.
This gives me
\[(x^3 - 2)^2 = 5^4.\]
Taking the square root of both sides, I get
\[x^3 - 2 = \pm 5^2.\]
Since I only want the positive value of $x$, I can ignore the negative case and focus on
\[x^3 - 2 = 5^2.\]
Adding $2$ to both sides and taking the cube root, I get
\[x = \sqrt[3]{5^2 + 2}.\]