Ae^kx: Find f(8) if f(x) = Ae^kx and f(2) = 10

[hereIgo]

This type of problem is new to me.

Find f(8) if f(x) = Ae^kx and f(2) = 10

OK, I realize that A must be a constant, and f(2) = Ae^2k = 10, but I'm unsure of how to go about finding k and A.

 

[galactus]

It appears the idea is to find k first and f(8) will be in terms of A.

If you have \(\displaystyle \L\\Ae^{2k}=10\), then solve for k.

\(\displaystyle \L\\e^{2k}=\frac{10}{A}\)

\(\displaystyle \L\\2k=ln(\frac{10}{A})\)

\(\displaystyle \L\\k=\frac{ln(\frac{10}{A})}{2}\)

Now, can you find what f(8) is?.

 

[daon]

Here's a "fun" way to figure it out: (although Galactus' seems more simple):

Notice:
$\displaystyle f(a+b) = Ae^{ak}e^{bk} = \frac{1}{A}f(a)f(b)$

Then:
\(\displaystyle f(8) = f(4+4) = \frac{1}{A}f(4)f(4) = \frac{1}{A}f(2+2)f(2+2) \\ = \frac{1}{A} \( \frac{1}{A}f(2)f(2) + \frac{1}{A}f(2)f(2) \) = \frac{2}{A^2} $f(2)$^2\)

So,
\(\displaystyle f(2)=10 \,\, \Rightarrow \,\, f(8) = \frac{2}{A^2} \( f(2) \)^2 = \frac{200}{A^2}\)

Also,
$\displaystyle f(8) = Ae^{8k}= \frac{200}{A^2} \,\, \Rightarrow \,\, A^3e^{8k} = 200$

Or for convenience sake, \(\displaystyle A^3e^{8k} = A^3\(e^{2k}\)^4 = $Ae^{2k}$^3e^{2k}=200.\)

Therefore we have a system of equations:
1. $\displaystyle \,\, Ae^{2k}=10$
2. \(\displaystyle \,\, \(Ae^{2k}\)^3e^{2k}=200\)

Substituting (1) into (2) gives an easy method for solving.

edit: For added fun (yeah, right): See if you can find f(8) without knowing what k is.