In a special case, we bring over the x^3 first because taking lg immediately runs into trouble.

lg y = lg [x^3 + p^(x - q)]

We can't split the lg on the right hand side because no laws of logarithms exist for that.

So, we bring over the x^3 first.

y - x^3 = p^(x - q)

Now we take logs on both sides.

lg [y - x^3] = lg p^(x - q)
lg [y - x^3] = (x - q) lg p
lg [y - x^3] = (lg p) x - q lg p

This is considered a linear form Y = mX + c because the values of Y (which is basically lg [y - x^3]) can be computed from raw data values of x and y, so that we are table to plot the data values of Y (lg [y - x^3]) against X (x) on a graph paper.

So if a set of data values is like x = 1 and y = 11, we can compute X and Y easily.

X = x = 1
Y = lg [y - x^3] = lg [11 - 1^3] = lg 10 = 1.00

If x = 2 and y = 108,

X = x = 2
Y = lg [y - x^3] = lg [108 - 2^3] = lg 100 = 2.00

Then, we are able to plot (1, 1.00) and (2, 2.00) on a graph paper to draw a straight line to find the gradient of the line.

This is just a rough idea of what's going on.

So in other words, your expressions for X and Y must be computable from raw data values of x and y. And there must be no x or y in the constant term.

In the original equation, it appears like Y = y and X = x^3. However, the constant term contains a variable x, and therefore cannot fit the form.

im still confused about the x^3+p^(x-q) issue

The key idea is that we will reach a dead end if apply lg immediately.

lg y = lg [x^3 + p^(x - q)]

We cannot proceed any further along this line, and therefore there is no way we can call out a linear form in this path.

In contrast,

y - x^3 = p^(x - q)

is worth considering. Note that I can even consider terms like y - x^3 as a single expression Y, because Y can contain a mix of x and/or y and I can compute y - x^3 from x and y alone.

Moreover, in this new form, I can apply lg to eventually get

lg [y - x^3] = (lg p) x - q lg p

from the expression.

Fortunately, the given starting question is unlikely to appear in the O Levels. It's more of just a practice for conversion.

Remember that in the form Y = mX + c,

- X and Y must contain a combinations of x and/or y only, or at least, the data values of X and Y can be obtained/evaluated from raw data values of x and y.

- m and c must not contain any x or y including in the power.

thx :)