To find the point of intersection of the graphs of \( y = \log_5 x \) and \( y = \log_{0.5} x \), we need to set the two equations equal to each other:
\[ \log_5 x = \log_{0.5} x \]
We can use the change of base formula to rewrite both logarithms in terms of base \( 10 \) (or any other base):
\[ \log_5 x = \frac{\log_{10} x}{\log_{10} 5} \] \[ \log_{0.5} x = \frac{\log_{10} x}{\log_{10} 0.5} \]
Thus, we can rewrite the equation:
\[ \frac{\log_{10} x}{\log_{10} 5} = \frac{\log_{10} x}{\log_{10} 0.5} \]
We can multiply both sides of the equation by \( \log_{10} 5 \cdot \log_{10} 0.5 \) (assuming \( \log_{10} x \neq 0 \)) to get:
\[ \log_{10} x \cdot \log_{10} 0.5 = \log_{10} x \cdot \log_{10} 5 \]
Now, if \( \log_{10} x \neq 0 \), we can divide both sides by \( \log_{10} x \):
\[ \log_{10} 0.5 = \log_{10} 5 \]
However, since \( \log_{10} 0.5 \) is actually negative, we need to check the case when \( \log_{10} x = 0 \). This occurs when:
\[ x = 1 \]
Now, let's evaluate both logarithms to see if they equal at this point:
\[ y = \log_5(1) = 0 \] \[ y = \log_{0.5}(1) = 0 \]
Thus, both equations give \( y = 0 \) when \( x = 1 \).
Therefore, the graphs of \( y = \log_5 x \) and \( y = \log_{0.5} x \) intersect at:
\[ (1, 0) \]
So, the answer is \( \boxed{(1, 0)} \).