linear and exponential graphs and x,y intercepts

I have two lines one linear (y=mx+c) and one exponential (y=c^bx) and
they intercept each other.  I believe it is possible to calculate where
the lines intercept.  Although I can do this by graphing the data, it
takes time and allows for user error so a calculation would speed
things along greatly.

The problem:

So for the first line using the equation y=mx+c the numbers for the
13kPa line are y=0.0021x+9e-16

And for the exponential line (y=c^bx) which we produce to using the
myograph is y=0.0022^0.0046x   (^=Exp)

I know that the two lines intercept at 1613.  But no matter what I try
I can't get the equations to simplify and produce a x value.

0
Mozam.ali (2)
7/31/2006 1:22:24 PM
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I would suggest that y=0.0021x+9e-16 is essentially y=0.0021x (the error in 
m is greater then the value of c)
This makes the math trivial
best wishes
-- 
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

<Mozam.ali@asterand.com> wrote in message 
news:1154352143.912290.52530@75g2000cwc.googlegroups.com...
>I have two lines one linear (y=mx+c) and one exponential (y=c^bx) and
> they intercept each other.  I believe it is possible to calculate where
> the lines intercept.  Although I can do this by graphing the data, it
> takes time and allows for user error so a calculation would speed
> things along greatly.
>
> The problem:
>
> So for the first line using the equation y=mx+c the numbers for the
> 13kPa line are y=0.0021x+9e-16
>
> And for the exponential line (y=c^bx) which we produce to using the
> myograph is y=0.0022^0.0046x   (^=Exp)
>
> I know that the two lines intercept at 1613.  But no matter what I try
> I can't get the equations to simplify and produce a x value.
> 


0
bliengme5824 (3040)
7/31/2006 2:46:48 PM
The way to solve this is to set the 2 equations equal to one another and 
then solve for x.  So, you will get
0.0021x+9e-16 = 0.0022*EXP(0.0046x)

For all practical purposes 9e-16 is zero.  So, we get
0.0021x = 0.0022*EXP(0.0046x)

This cannot be solved algebriacally but it can be solved numerically.

Designate a cell, say B2 as the one containing the X value.  Then, in C2, 
enter the formula =0.0021*B2.  In D2, enter the formula
=0.0022*EXP(0.0046*B2).  In E2 enter the formula =C2-D2.

Now, use GoalSeek (Tools | Goal Seek...) to set E2 to zero by changing B2.  
Though, given the scale of the numbers, I would go with Solver (Tools | 
Solver...)  Set the target as E2, set the desired value of 0 and click 
Solve.

I get two results, both borne out by a chart.  The first intersection is 
reached if you start with an initial value of 0 in B2.  The second by 
starting with an initial value of 2000.

And, for the larger result, I don't get 1631 but 1593.

-- 
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

In article <1154352143.912290.52530@75g2000cwc.googlegroups.com>, 
Mozam.ali@asterand.com says...
> I have two lines one linear (y=mx+c) and one exponential (y=c^bx) and
> they intercept each other.  I believe it is possible to calculate where
> the lines intercept.  Although I can do this by graphing the data, it
> takes time and allows for user error so a calculation would speed
> things along greatly.
> 
> The problem:
> 
> So for the first line using the equation y=mx+c the numbers for the
> 13kPa line are y=0.0021x+9e-16
> 
> And for the exponential line (y=c^bx) which we produce to using the
> myograph is y=0.0022^0.0046x   (^=Exp)
> 
> I know that the two lines intercept at 1613.  But no matter what I try
> I can't get the equations to simplify and produce a x value.
> 
> 
0
7/31/2006 3:11:13 PM
More help needed.

I do not have solver installed on my PC and the logistics of getting it
put onto every PC at work is going to be impossible.  So I can't use
the solver.  I do have GoalSeak if thats any help.

Next although the y-value for this equation is near zero that will not
be the case in all the eqautions i deal with.  the equations are just
an example.

I want to know at what X-value do the two lines cross.

I would preferably want it as an eqaution because I need to construct
it using XL.  Also I have done much in the way of mathematics since uni
(about 7 years ago) so you will have to go slow.  I know it's a
terrible excuse, but I need to understand it all.

mc+c=c^bx   (^=exp).

0
Mozam.ali (2)
8/7/2006 1:29:18 PM
Reply:

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