Best Answer

y = c ekx

passing through (1, 180) and (3, 20).

(Note: All of the 'log' in the following are natural logs.)

log(y) = log(c) + k log(x)

Log(180) = log(c) + k log(1) = log(c)

c = 180

log(20) = log(180) + k log(3)

log(20) - log(180) = k log(3)

k = log(20/180) / log(3) = - log(9) / log(3) = -2

y = 180 e-2xIn checking my work, I find that this doesn't work at all.

Oh woe! Where have I failed ?

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Here's what I did:

y=Cekx which passes through the points (1,180) & (3,20)

(1) Substitute the values of x & y to create two equations:

180 = Cek & 20 = Ce3k

(2) Rearrange in terms of the constant, C:

C=180/ek & C=20/e3k

(3) Since both sets of points satisfy the exponential function, the constant will be the same so:

180/ek = 20/e3k

(4) Now rearrange and solve for k:

180e3k = 20ek

9 = e(k-3k)

ln(9) = ln(e-2k)

ln(9) = -2k

k = -ln(9)/2 which is approx. -1.09861

(5) Now substitute k into one of the equations to solve for C:

C =180/e(-1.09861)

C = 539.9987 or rounded to 540

So the exponential function that includeds the points (1,180) & (3,20) can be approximated as:

y = 540ekx , where k = -1.09861

Q: What is the exponential function in the form y equals c times e to the power of kx that passes through the points 1 180 and 3 20?

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