1. **Problem Statement:** We want to understand why the line $\log_8 y$ is a linear function of $\log_4 y$. 2. **Recall the change of base formula:** For any positive numbers $a,b,c$ with $a \neq 1$, $$\log_a b = \frac{\log_c b}{\log_c a}$$ This means we can express logarithms with different bases in terms of a common base. 3. **Express $\log_8 y$ in terms of $\log_4 y$: ** Using the change of base formula with base 4, $$\log_8 y = \frac{\log_4 y}{\log_4 8}$$ 4. **Calculate $\log_4 8$: ** Since $8 = 2^3$ and $4 = 2^2$, $$\log_4 8 = \log_4 (2^3) = 3 \log_4 2$$ But $\log_4 2 = \frac{1}{2}$ because $4^{1/2} = 2$, so $$\log_4 8 = 3 \times \frac{1}{2} = \frac{3}{2}$$ 5. **Substitute back:** $$\log_8 y = \frac{\log_4 y}{3/2} = \frac{2}{3} \log_4 y$$ 6. **Interpretation:** This shows $\log_8 y$ is exactly $\frac{2}{3}$ times $\log_4 y$, which is a linear relationship of the form $$y = mx$$ where $m = \frac{2}{3}$. **Final answer:** $$\log_8 y = \frac{2}{3} \log_4 y$$ This means the graph of $\log_8 y$ versus $\log_4 y$ is a straight line through the origin with slope $\frac{2}{3}$.