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~~~~~~~~Need basic equation help~~~~~~~~

zdubstepx

zdubstepx

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hey sythers i need help with my dumb math project..

so basicaly i have a table of points and a graph made but i have no idea how to find the equation.

i will post the data table here:

YEARS | mW
1982 | 7.0
1983 | 13.0
1984 | 10.0
1985 | 6.0
1986 | 6.6
1987 | 6.9
1988 | 10.0
1989 | 13.0
1990 | 13.8
1991 | 14.9
1992 | 15.6
1993 | 21.0
1994 | 26.1
1995 | 31.1
1996 | 35.5
1997 | 46.4
1998 | 50.6
1999 | 76.9


and the graph here:

graph.png



Please if you know how to find the equation of this graph please post the equation here! thanks!!!
 
yes ik its a exponential curve thanks. i need the equation of the line if you can get that for me or tell me how to get that using a TI instrument graphing calc i would appreciate it so i can get back to pwning kids on rs.
 
On a TI 83 or better you can plot it and use the exponential regression to find the line. It's been awhile but you might be able to use some tutorial to do it.

Although excel would be easier.
 
if anyone can just input my data in excel or something and gimmy a equation i would be really happy. lol
 
zdubstepx said:
yes ik its a exponential curve thanks. i need the equation of the line if you can get that for me or tell me how to get that using a TI instrument graphing calc i would appreciate it so i can get back to pwning kids on rs.
Click to expand...

In the TI graphing calculator there is an option to add X and Y data to a graph (I believe it's the graph button, never used a calculator for graphing). Lets see if I can just do this in my head:

Y = Yo*e^kt Where k is constant and t is changing (can be x in this case)
will be the form. Solving for K using the last data point:

(76.9 / 7) =
11 = e^(k*17)
ln(11) = k*17
ln(11)/17 = k

Now that we have a k value, plug it back into the original equation

Y = 7 * e^(k*t)

REMEMBER to keep your t ranging from 0-19 in steps of one (0 being 1982, 17 being 1999)


Did that help or should I explain further what I did?

(throw out large, erroneous data values, like the second point)
 
y = 5.203e0.1407x
 
e is exponential, not e in scientific notation. I think you should be doing your homework more if you mistook that for scientific notation.

y = 5.203 * e ^(0.1407*x)
e = ~ 2.7
 
YES!!!! thank you!!! but how would i write that equation so my math teacher dont think i coped it off the web? lol.
 
Just tell your professor you put the numbers into excel and graphed them using a scatter plot, and then added a trendline. The instructions for doing that were all online.

If you need to explain to her how you solved the equation, replace x with t and follow my steps a few posts above when I solved for k. plug k back in and you'll have something near the equation I gave you.

Be sure to round the equation I gave you as well. Make it look simple and crude. Those decimal places don't mean anything, round them.
 
yeah cuz i gotta explain how each part of that equation changes the graph from the perent function of ab^x. so i would just say 5.203 * e ^(0.1407*x) but i would need to explain each part of it out.
 
Hmm. Not quite sure I get what you mean.

Use this equation: Y = yo*e^(k*x) where k is a constant and x is your x value.
Solving for the x is easy and y. your x axis goes from 0 - 17 in increments of 1, and you're given the y values with those x's. To solve for the Yo value, set e^(constant*t) to 1, so you have the equation:
Y = Yo. And since the first Y is at 7 when x=0, you can say Yo is 7.

Throw out the second data point, it's garbage. (you're allowed to do that when you have erroneous data)

To solve for the constant (k), you do this:

Y = 7 * e^(k*x)

Lets use the last data value since it's the best example of the exponential growth:

(17,76.9) is our (x,y)

76.9 = 7 * e^(k*17)

Divide both sides by 7 to get that e alone.
77 / 7 = e^(k*17)
11 = e^(k*17)

If you ever need to get rid of an e (in this case we need to get rid of the e in order to solve for the k that is part of the exponent of e) we can multiply both sides by ln (natural log).

ln(11) = ln(e^(k*17))

ln(11) = k*17
Now to get k alone, divide by 17.

ln(11)/17 = k.

Now that we have a number for k, plug it back into the original equation: y = 7*e^(k*x) leaving y and x as variables.
 
malyce said:
Hmm. Not quite sure I get what you mean.

Use this equation: Y = yo*e^(k*x) where k is a constant and x is your x value.
Solving for the x is easy and y. your x axis goes from 0 - 17 in increments of 1, and you're given the y values with those x's. To solve for the Yo value, set e^(constant*t) to 1, so you have the equation:
Y = Yo. And since the first Y is at 7 when x=0, you can say Yo is 7.

Throw out the second data point, it's garbage. (you're allowed to do that when you have erroneous data)

To solve for the constant (k), you do this:

Y = 7 * e^(k*x)

Lets use the last data value since it's the best example of the exponential growth:

(17,76.9) is our (x,y)

76.9 = 7 * e^(k*17)

Divide both sides by 7 to get that e alone.
77 / 7 = e^(k*17)
11 = e^(k*17)

If you ever need to get rid of an e (in this case we need to get rid of the e in order to solve for the k that is part of the exponent of e) we can multiply both sides by ln (natural log).

ln(11) = ln(e^(k*17))

ln(11) = k*17
Now to get k alone, divide by 17.

ln(11)/17 = k.

Now that we have a number for k, plug it back into the original equation: y = 7*e^(k*x) leaving y and x as variables.
Click to expand...




thanks again man really helped!
 
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