help! how do you make the points (3,4) & (7,64) into y=ke^cx?! please help?

Question

and also into y=ab^x


I REALLY NEED HELP< PLEASE!!!

Answer 1

plug in for x and y and solve the systems...

64 = ke^7c
4 = ke^3c

divide the equations and the k's cancel out and you get:
16 = e^4c
ln 16 = 4c
c = (ln 16)/ 4
c = ln 16^1/4
c = ln 2

4 = ke^(3*(ln 2)
4 = ke^(ln 2^3)
4 = k*8
k = 1/2
********
64 = a*b^7
4 = a*b^3

Divide the equations and the a's cancel out and you get:

16 = b^4
b=2
4 = a*2^3
4 = 8a
1/2 = a

Answer 2

y = ke^cx

take the natural log of both sides

ln y = cx + ln k.

Since the unknowns here are c and k, not x and y, I'll rearrange:

ln k = ln y - cx

Now we can write a system of linear equations:

ln k = ln 4 - 3c,
ln k = ln 64 - 7c.

Subtracting the two equations:

0 = 4c + ln 4 - ln 64 = 4c + ln (1/16).

Solving,

c = - (1/4) ln (1/16) (Q.et D.)

Then, back substituting, and taking e to the power of both sides:

ln k = ln 4 + (3/4) ln (1/16)

k = 4e^{ (3/4) ln (1/16) } (exact)

Does that make sense? You can use the same basic method of simultaneous equations for y = ax + b.....

Hope that answers your question,
~W.O.M.B.A.T.

"Logic! Why don't they teach logic in these schools!?"
-C.S. Lewis

Answer 3

You can get c by taking the ratio because k will factor out. y(x=7)/y(x=3) = 64/4 = 16 = (ke^7c)/(ke^3c) = e^(7c - 3c) = e^4c. Taking the natural log of both sides, we get: Ln(16) = 4c or c = Ln(16)/4 = Ln(2^4)/4 = 4Ln(2)/4 = Ln(2).

Once you have c, you can use either point to get k. For example, 4 = k e^(ln(2)*3) = ke^ln(2^3) = ke^ln(8) = 8k or k = 4/8 = 1/2.

The final equation is therefore y = 0.5e^(ln(2)*x) = 2^(x-1). Can you figure out why? This also answers your other question.

Math Rules!

Answer 4

vejamos carlos:
y=ke^cx

64=ke^c7
------------ ==> 16 = e^(7c - 3c) ===> 2^4 = e^(4c)
4 = ke^c3

ln 2^4 = ln e^(4c)
4 ln 2 = (4c) ln e
4 ln 2 = 4c
c = ln 2

y = ke^(x ln2)

Answer 5

Best way to do it would be to input those points into a calculator or spreadsheet and calculate the exponential regression.