# Express the following equations in a form suitable for drawing a straight line graph

• Jun 30th 2012, 11:49 PM
fActor
Express the following equations in a form suitable for drawing a straight line graph
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)
• Jul 1st 2012, 01:59 AM
BobP
Re: Express the following equations in a form suitable for drawing a straight line gr
Assuming that a and b are unknown constants to be determined ?

For the first relationship, take logs

$\ln(y)=ax+by,$ and then divide by $y,$ (you could also divide by $x$, that would lead to an alternative equation),

$\frac{\ln(y)}{y}=a\left(\frac{x}{y}\right)+b.$

Then construct the new variables $Y=\ln(y)/y$ and $X=x/y.$

That will give you the straight line $Y=aX+b$ which has a slope of $a$ and an intercept of $b.$

You can do a similar thing with the second example.
• Jul 8th 2012, 03:52 AM
fActor
Re: Express the following equations in a form suitable for drawing a straight line gr
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?
• Jul 9th 2012, 01:57 AM
BobP
Re: Express the following equations in a form suitable for drawing a straight line gr
Yes, that should work.