1. The problem involves understanding logarithmic relationships and solving for variables given certain logarithmic values. 2. We use the logarithm properties and linear equations. Recall that $\log_{10} y$ means the base-10 logarithm of $y$, and to find $y$ from $\log_{10} y = a$, we use $y = 10^a$. 3. Given $y = 4.8$, calculate $\log_{10} 4.8 = 0.681$. 4. Calculate slope $m$ using points $(1.5, 0.4)$ and $(4.8, 1.6)$: $$m = \frac{1.6 - 0.4}{4.8 - 1.5} = \frac{1.2}{3.3} = 0.3636$$ 5. The equation of the line is $y = mx + c$. Using $m=0.3636$ and point $(1.5, 0.4)$: $$0.4 = 0.3636 \times 1.5 + c \Rightarrow c = 0.4 - 0.5454 = -0.1454$$ 6. So the line equation is: $$y = 0.3636 n - 0.1454$$ 7. Given $\log_{10} y = 0.681$, substitute $y$: $$0.681 = 0.3636 n - 0.1454$$ 8. Solve for $n$: $$0.3636 n = 0.681 + 0.1454 = 0.8264$$ $$n = \frac{0.8264}{0.3636} = 2.273$$ 9. For $\log_{10} n = 6.28$, find $n$: $$n = 10^{6.28} = 1.91 \times 10^{6}$$ 10. For $\log_{10} k = 0.08$, find $k$: $$k = 10^{0.08} = 1.202$$ Final answers: - Slope $m = 0.3636$ - Intercept $c = -0.1454$ - $n$ when $\log_{10} y = 0.681$ is approximately $2.273$ - $n$ when $\log_{10} n = 6.28$ is approximately $1.91 \times 10^{6}$ - $k$ when $\log_{10} k = 0.08$ is approximately $1.202$