Search over 10,000 free study notes

First, we need to re-write it as e to the power something. The definition of log base e is that e^log(y) = y. We can put our expression into this equality too. So a^x = e^log(a^x), so we use log rules to ...

So in order to derive the quadratic equation we need to start with a quadratic that can represent any quadratic equation. Lets start with ax²+bx+c=0.

We will use the completing the squar...

So we have y = 0 and therefore x² - 6x - 16 =0 

This is then factorised to (x+2)(x-8) = 0 and therefore we have x = -2, x = -8

To find the gradient we need to find dy/dx

So...

For the most part, this comes down to practice. The insight in seeing which identity gives you the most efficient path to the solution comes with experience, however a little trick I like to use (silly as...

NB.: Treat all log as log2 for purpose of formatting log(x) - log(z) = log(x/z) alog(b) = log(b^a) log((11y - 3)/3) - log(y^2) = 1 log((11y - 3)/3y^2) = 1 11y - 3 / 3y^2 = 2^1 11y - 3 = 6y^2 6y^2 - 11y + ...