Quick questions on The linear law explained: O-Level A-Maths

Not every relationship is a clean power or exponential law, so the general skill is to rearrange any equation into the shape $Y = mX + c$ and read off what $Y$ and $X$ must be. For $y = ax^2 + b$, no logarithms are needed: plotting $y$ against $x^2$ is already linear, with gradient $a$ and intercept $b$. For $\tfrac{1}{y} = ax + b$, plot $\tfrac{1}{y}$ against $x$. The method is to isolate the part containing the unknown constants as a linear combination of two computable quantities, then those two quantities become your axes.

With real experimental points, the plotted data will not lie perfectly on a line, so you draw a line of best fit and take its gradient and intercept from two well-separated points on that line, not from raw data points. Using points far apart on the fitted line reduces the effect of reading errors on the gradient. Then convert as usual, for instance $a = 10^{\text{intercept}}$. Emphasising the best-fit line, rather than any single data point, is what makes the constants reliable, and it is the practical reason the linear law is so useful for analysing measurements.

Use the same base throughout; if you take $\ln$, then the intercept is $\ln a$ and $a = e^{\text{intercept}}$.

Write $y = ax^n$ in straight-line form and state what is plotted on each axis. [2 marks]

A plot of $\lg y$ against $x$ for $y = Ab^x$ has gradient $0.5$. Find $b$. [2 marks]

For $y = ax^n$, the $\lg y$-against-$\lg x$ line has intercept $2$. Find $a$. [2 marks]