Let's go through each part of the question step-by-step, focusing on the additional explanation for part (b) once we've addressed part (a).

Part (a): Finding the equation linking $\log_{10} y$ with $\log_{10} x$

1. Identify the given points: The line passes through the points $(0, 4)$ and $(6, 0)$ on the coordinate system where the axes are labeled $\log_{10} y$ (vertical) and $\log_{10} x$ (horizontal).

2. Calculate the slope (gradient): The slope $m$ of a line through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Using the points $(0, 4)$ and $(6, 0)$: $m = \frac{0 - 4}{6 - 0} = \frac{-4}{6} = -\frac{2}{3}$

3. Use the point-slope form of the line equation: We can write the line equation in the form: $y - y_1 = m(x - x_1)$ Substituting $m = -\frac{2}{3}$ and using point $(0, 4)$ as $(x_1, y_1)$: $\log_{10} y - 4 = -\frac{2}{3}(\log_{10} x - 0)$ Simplifying this: $\log_{10} y = -\frac{2}{3} \log_{10} x + 4$

   Thus, the equation linking $\log_{10} y$ and $\log_{10} x$ is: $\log_{10} y = -\frac{2}{3} \log_{10} x + 4$

Part (b): Express $y$ in the form $y = px^q$

To express $y$ in terms of $x$ using the form $y = px^q$, we can interpret the equation from part (a) in terms of exponentials.

1. Rewrite the equation in exponential form: We have: $\log_{10} y = -\frac{2}{3} \log_{10} x + 4$ We can rewrite this as: $\log_{10} y = \log_{10} \left( x^{-\frac{2}{3}} \right) + \log_{10} (10^4)$ Using the property $\log_{10} (a) + \log_{10} (b) = \log_{10} (a \cdot b)$: $\log_{10} y = \log_{10} \left(10^4 \cdot x^{-\frac{2}{3}} \right)$

2. Rewrite in terms of $y$: Since $\log_{10} y = \log_{10} \left(10^4 \cdot x^{-\frac{2}{3}} \right)$, we can conclude: $y = 10^4 \cdot x^{-\frac{2}{3}}$

3. Express in the form $y = px^q$: Comparing $y = 10^4 \cdot x^{-\frac{2}{3}}$ with $y = px^q$, we see that:

   • $p = 10^4$
   • $q = -\frac{2}{3}$

   So, the final answer for part (b) is: $y = 10^4 \cdot x^{-\frac{2}{3}}$

Summary

The solutions are:

• (a) The equation linking $\log_{10} y$ with $\log_{10} x$ is: $\log_{10} y = -\frac{2}{3} \log_{10} x + 4$
• (b) In the form $y = px^q$, we have: $y = 10^4 \cdot x^{-\frac{2}{3}}$ where $p = 10^4$ and $q = -\frac{2}{3}$.

Would you like more details on any step?

Related Questions

1. How would the equation change if the slope were different?
2. What would happen to the form $y = px^q$ if the intercept were different?
3. How does the logarithmic form help in linearizing exponential relationships?
4. Can we generalize this approach to other bases besides 10?
5. How does changing the base of the logarithm affect the values of $p$ and $q$?

Tip

In problems involving logarithmic relationships, try to express everything in terms of exponentials if you need to find a power-law relationship. This makes the transformation straightforward.