1. **Stating the problem:** We have a table with values for variables $x$ and $y$: $$x = 2, 4, 16$$ $$y = 4, 8, 32$$ A graph of a straight line is drawn with the $y$-axis as $\log_{p+1} y$ and the $x$-axis as $\log_{p+1} x$. We need to find the values of $p$ and $q$ assuming the line equation is of the form: $$\log_{p+1} y = q \log_{p+1} x$$ 2. **Formula and rules:** Recall the logarithm power rule: $$\log_a b^c = c \log_a b$$ If the graph is a straight line through the origin, then $y = x^q$ in logarithmic form: $$\log_{p+1} y = q \log_{p+1} x$$ 3. **Using the data points:** Calculate $\log_{p+1} y$ and $\log_{p+1} x$ for each pair $(x,y)$: From the equation: $$\log_{p+1} y = q \log_{p+1} x$$ Rewrite $y = x^q$ in terms of base $p+1$ logarithms. 4. **Expressing $y$ in terms of $x$ and $q$:** Given $y = x^q$, check if this fits the data: - For $x=2$, $y=4$ implies $4 = 2^q$ so $q = \log_2 4 = 2$ - For $x=4$, $y=8$ implies $8 = 4^q$ so $q = \log_4 8$ Calculate $\log_4 8$: $$\log_4 8 = \frac{\log 8}{\log 4} = \frac{\log 2^3}{\log 2^2} = \frac{3 \log 2}{2 \log 2} = \frac{3}{2} = 1.5$$ - For $x=16$, $y=32$ implies $32 = 16^q$ so $q = \log_{16} 32$ Calculate $\log_{16} 32$: $$\log_{16} 32 = \frac{\log 32}{\log 16} = \frac{\log 2^5}{\log 2^4} = \frac{5 \log 2}{4 \log 2} = \frac{5}{4} = 1.25$$ 5. **Observing $q$ values:** The $q$ values are not consistent, so the assumption $y = x^q$ with base 2 logarithm is not fitting all points. 6. **Using the given logarithm base $p+1$:** We want to find $p$ such that the points lie on a straight line: $$\log_{p+1} y = q \log_{p+1} x$$ Rewrite as: $$\frac{\log y}{\log (p+1)} = q \frac{\log x}{\log (p+1)}$$ Simplifies to: $$\log y = q \log x$$ which means $y = x^q$ regardless of $p$. 7. **Finding $q$ from the data:** Calculate $q$ using pairs: Between $(2,4)$ and $(4,8)$: $$q = \frac{\log y_2 - \log y_1}{\log x_2 - \log x_1} = \frac{\log 8 - \log 4}{\log 4 - \log 2} = \frac{\log 2^3 - \log 2^2}{\log 2^2 - \log 2} = \frac{3 \log 2 - 2 \log 2}{2 \log 2 - \log 2} = \frac{\log 2}{\log 2} = 1$$ Between $(4,8)$ and $(16,32)$: $$q = \frac{\log 32 - \log 8}{\log 16 - \log 4} = \frac{5 \log 2 - 3 \log 2}{4 \log 2 - 2 \log 2} = \frac{2 \log 2}{2 \log 2} = 1$$ So $q=1$. 8. **Finding $p$:** Since the graph is a straight line with slope $q=1$ in base $p+1$ logarithms, check the intercepts. Using the point $(x=2, y=4)$: $$\log_{p+1} 4 = q \log_{p+1} 2$$ Since $q=1$: $$\log_{p+1} 4 = \log_{p+1} 2$$ But $\log_{p+1} 4 = 2 \log_{p+1} 2$ (because $4=2^2$), so this is a contradiction unless $\log_{p+1} 2 = 0$ which is impossible. 9. **Re-examining the problem:** The problem states the graph is a straight line with $y$-axis $\log_{p+1} y$ and $x$-axis $\log_{p+1} x$, so the line equation is: $$\log_{p+1} y = q \log_{p+1} x + c$$ where $c$ is the intercept. Using two points to find $q$ and $c$: For $(2,4)$: $$\log_{p+1} 4 = q \log_{p+1} 2 + c$$ For $(4,8)$: $$\log_{p+1} 8 = q \log_{p+1} 4 + c$$ Subtracting: $$\log_{p+1} 8 - \log_{p+1} 4 = q (\log_{p+1} 4 - \log_{p+1} 2)$$ Using log subtraction: $$\log_{p+1} \frac{8}{4} = q \log_{p+1} \frac{4}{2}$$ $$\log_{p+1} 2 = q \log_{p+1} 2$$ So $q=1$. 10. **Finding $p$ using the point $(2,4)$:** $$\log_{p+1} 4 = 1 \times \log_{p+1} 2 + c$$ $$c = \log_{p+1} 4 - \log_{p+1} 2 = \log_{p+1} \frac{4}{2} = \log_{p+1} 2$$ 11. **Using the point $(16,32)$:** $$\log_{p+1} 32 = 1 \times \log_{p+1} 16 + c$$ Substitute $c$: $$\log_{p+1} 32 = \log_{p+1} 16 + \log_{p+1} 2 = \log_{p+1} (16 \times 2) = \log_{p+1} 32$$ This is consistent. 12. **Conclusion:** The slope $q=1$ and intercept $c = \log_{p+1} 2$. Since $c$ is a constant, the problem likely wants $p$ such that the base $p+1$ satisfies the logarithm properties. Choose $p+1 = 2$ (base 2 logarithm) for simplicity: Then $p = 1$. **Final answers:** $$p = 1$$ $$q = 1$$