Looking for some help in reducing an equation to linear form, y=1/(x-a)(x-b), a and b are unknowns.?

Question

I've taken the logs:
log y = log 1 - log (x-a)(x-b)
so
log y = -log(x-a)(x-b)
so
log y = -log(x-a) - log(x-b)
not sure if this is correct so far, if yes, still can't see how it fits into y =mx+c
Please help!!!

Answer 1

y = 1/(x-a)(x-b)
y = 1/(x^2 - bx - ax + ab)
y = 1/(x^2 - (a+b)x + ab)

This equation cannot be reduced to a linear equation as it is not linear. The denominator consists of a quadratic equation and it is also a reciprocal.

Answer 2

You won't be able to reduce this equation to y=mx+b, because the equation is not linear, it's quadratic.

Answer 3

Until here it is correct:

log y = log 1 - log (x-a)(x-b)
so
log y = -log(x-a)(x-b)
so
log y = -log(x-a) - log(x-b)

But here is the final reduction possible!!!