Le t f(x) be an exponential function?

Question

with f(7)= 3 and f(13)= 24. What is f(19)?

As always, your help is much appreciated!

Answer 1

Massive edit: Okay, when I first did the problem, I implicitly assumed that this was a function of the form f(x)=a^x for some a, so that it would obey the property f(x+y)=f(x)*f(y) and f(x-y)=f(x)/f(y). I then used this flawed assumption to produce the "derivation"

f(19)
f(6+13)
f(13-7+13)
f(13+13-7)
f(13+13)/f(7)
f(13)²/f(7)
24²/3
f(19)=192

Thus accidentally obtaining the right answer. Of course, as later posters have noted, what your teacher most likely means by an exponential function here is ca^x, not a^x (indeed, it is impossible for c=1). The fact that I got the right answer is not entirely a fluke of chance, since:

f(13)²/f(7) = (ca^13)²/(ca^7) = c²a^26/(ca^7) = ca^19 = f(19)

So this would have produced the right answer regardless of what values had been given for f(7) and f(13). Nonetheless, the derivation was logically flawed - sabseg50 was the first to produce a logically correct answer. I leave this note up here for the purpose of enlightenment and so that people will not be left wondering "hey, where did Pascal's answer go?"

Answer 2

19 - 13 = 6 = 13 - 7.
Since f is exponential, f(19) / f(13) = f(13) / f(7). So f(19) = 24 * (24/3) = 24 * 8 = 192.

[After noticing Pascal's edit]:
Actually, Pascal, this answer is also logically correct, since it was in fact based on the general assumption that the function is of form y = c.e^kx. It relies only on a constant difference in the argument translating into a constant ratio for the result, which is true for any such expression (since y2 / y1 = e^kx2 / e^kx1 = e^k(x2-x1).) I probably should have explained this point, but the intent of my answer was to point out the shortcut available due to the fact that the difference between the two known values was the same as that between one of the known values and one of the unknown values.

Answer 3

This is a longer way of doing it, but produces not only the value of f(19) but also the general expression of the function.

Let f(x) = A e^(Bx) , where A and B are unknown constants

Since 3 = f(7) = Ae^(B7) and 24 = f(13) = Ae^(B13),

8 = 24/3 = f(13)/f(7) = Ae^(B13)/Ae^(B7) = e^(B6)

So, applying natural logatihms we obtain

ln(8) = ln( e^(B6) ) = B6 since ln(e^(Y)) = Y for all Y

Solving for B,

B = (ln(8))/6 = (ln(2^3))/6 = ln(2^(3/6)) = ln(2^(1/2))

Since e^(ln(Z)) = Z for all Z > 0,

e^(Bx) = e^(xln(2^(1/2)) = e^(ln(2^(x/2)) = 2^(x/2)

So far we have

f(x) = A2^(x/2)

Therefore

3 = f(7) = A2^(7/2)

Solving for A:

A = 3/2^(7/2)

Finally,

f(x) = ( 3/2^(7/2) ) 2^(x/2) or

f(x) = 3 * 2^( (x-7)/2 ) since 2^m / 2^n = 2^(m-n)

In particular,

f(19) = 3 * 2^( (19-7)/2 ) = 3 * 2^(12/2) = 3 * 2^6 = 3 * 64 = 192

Answer 4

ae^7b = 3
ae^13b = 24
(ae^7b)(ae^6b) = 24
(ae^6b = 8
(ae^19b) = (ae^13b)(ae^6b)
(ae^19b) = 24*8 = 192