An easy log question? [modified]?

Question

Say I have two points: (x1,y1) = (0+,0+) close to zero, and (x2,y2). The second point is in the top right quadrant. (see fig). Under what conditions can I find p such that e^p goes through both points and looks something like the curved line I have drawn?

Thanks!

http://s153.photobucket.com/albums/s235/s1020099/?action=view¤t=expo-1.jpg

sorry, e^p should be x^p

Answer 1

A simple graph of e^p to go through this point would have to be y = e^p - 1 or e^p - ln(p + e) to go through (0,0) since e^p at p = 0 would be 1.

if you have MATLab, type the following lines and you will see the graph

x = 0:0.1:10;
y = exp(x) - log (x+exp(1));
plot (x,y,'b');

In MATLab, log is actually the natural logarithm, not log base 10. Now, to solve y = e^p - 1
http://img251.imageshack.us/img251/3840/graphoa0.jpg

y + 1 = e^p
ln (y + 1) = ln (e^p)

from log laws

log (a^b) = b log a, therefore

p ln (e) = ln (y + 1)
p = ln (y + 1)

That's all there is to it, if that's not what you're looking for, let me know and I can give you an alternate solution for what you need.

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If the graph is x^p then this is easier than the above.

y = x^p

ln y = p ln x (or log, your preference, using log laws from above still)

p = (ln y) / (ln x)

http://img338.imageshack.us/img338/5359/graphkv9.jpg

The first graph is the graph of the function y = x^p
The blue graph shows how x is varying over itself (in constant intervals), whereas the red graph shows how y is varying with changes in x (in an exponential manner)

Answer 2

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Answer 3

dont know conditions but let p=log(.) will get u back to quadratics(log to the base e) otherwise i have never seen such an exponential

Answer 4

y=a*e^x y1=ae^x1
y2=ae^x2
y2-y1= a(e^x2-e^x1) so a = (y2-y1)/(e^x2-e^x1)
This exponential passes through both points