Same story.

The equation would have been linear, had the unknown power b not been there.

Because we can't compute x^b from raw data values of x (unlike fixed, known powers like 2 or 3).

So, we take lg.

lg y = lg ax^b
lg y = lg a + lg x^b
lg y = b lg x + lg a

i.e. Y = mX + c

where we plot lg y (vertical) against lg x (horizontal).

Compute your lg y and lg x from raw data values of x and y. Then, plot a best fit line as usual.

The calculated gradient of the line is just nice the value of b.

The vertical intercept reading of the line is your lg a, so a = 10^(your vertical intercept).

We find the data point which lies far off the best fit line. We take x-value of the raw data (and hence the lg x value of the data point) to be correct, so we trace out the "correct" position of lg y from the graph.

Then, the correct reading y = 10^(your found reading).

btw for question like part(i), how do we answer them tho, i dont really know what they are asking

Basically what I wrote earlier, up to the plotting.

“Explain how a straight line graph may be drawn to represent the given equation...”

This is asking us to obtain an equation like

lg y = b lg x + lg a

which is a straight line equation where Y = lg y and X = lg x.

“...and draw it for the given data”

From the given raw data of x and y, we compute lg x and lg y before marking out their positions on the two axes. Horizontal axis is lg x. Vertical axis is lg y. Then we draw a best fit line.

thx :)