# Math Help - Linear Law

1. ## Linear Law

My attempt at manipulating the equation

$y^2=e^{ax+4}$
$2lny=(-ax+4)lne$
$lny=\frac{(-ax+4)}{2}$
$lny=\frac{-a}{2}x+2$

2. Originally Posted by Punch

My attempt at manipulating the equation

$y^2=e^{ax+4}$
$2lny=(-ax+4)lne$
$lny=\frac{(-ax+4)}{2}$
$lny=\frac{-a}{2}x+2$
hi

gradient $=-\frac{a}{2}=\frac{-4-b}{2}$ -- equation 1

the y-intercept is 2 , so we know another point (0,2)

gradient $=-\frac{a}{2}=\frac{b-2}{2}$ -- equation 2

so now you hv 2 equations , whats next ?

3. You are right

Using the point (4,-4) gives you the value for a
$-4 = -\frac{a}{2} \times 4 + 2$

Then using the point (2,b) gives you the value for b

4. From $lny=\frac{-a}{2}x+2$, substituting the point (4,-4)

I would have to take $lny=ln(-4)$ which shows error 2 in the calculator.
confused

5. Originally Posted by Punch
From $lny=\frac{-a}{2}x+2$, substituting the point (4,-4)

I would have to take $lny=ln(-4)$ which shows error 2 in the calculator.
confused
ln y (the whole thing) is -4 not the y inside the ln .

6. Oh... This means, $lny=\frac{-a}{2}x+2$ is expressed as Y=mX+C where where Y is the y coordinate and X is the x-coordinate

7. Originally Posted by Punch
Oh... This means, $lny=\frac{-a}{2}x+2$ is expressed as Y=mX+C where where Y is the y coordinate and X is the x-coordinate
yes