Express the following equations in a form suitable for drawing a straight line graph, stating its axes, gradient and vertical intercept.
1. y = e^(ax + by)
2. y = (a / b)^(xy + 1)
Express the following equations in a form suitable for drawing a straight line graph, stating its axes, gradient and vertical intercept.
1. y = e^(ax + by)
2. y = (a / b)^(xy + 1)
Assuming that a and b are unknown constants to be determined ?
For the first relationship, take logs
$\displaystyle \ln(y)=ax+by,$ and then divide by $\displaystyle y,$ (you could also divide by $\displaystyle x$, that would lead to an alternative equation),
$\displaystyle \frac{\ln(y)}{y}=a\left(\frac{x}{y}\right)+b.$
Then construct the new variables $\displaystyle Y=\ln(y)/y$ and $\displaystyle X=x/y.$
That will give you the straight line $\displaystyle Y=aX+b$ which has a slope of $\displaystyle a$ and an intercept of$\displaystyle b.$
You can do a similar thing with the second example.
For the second question, this is what I did:
lg y = (xy + 1)(lg a - lg b)
lg y = xy lg a - xy lg b + lg a - lg b
lg y = (lg a/b)xy + lg a/b
So the Y-axis = lg y, X-axis = xy, gradient = lg a/b and vertical intercept = lg a/b
Is this correct?