If f is an exponential function, and f(0) = 1,000 and f(1) = 500, find the defining equation for f.?

Answer 1

Since f is an exponential function of a variable, say 'x'. Our equation is in general:

f = A*(e^(B*x))

We have to find the constants A & B. So, we use the two given conditions:

f(0) = 1000, i.e. at x=0, f=1000

Thus,
1000 = A*(e^(B*0)) = A*1 = A
A = 1000

Now, our eqn has been simplified one step to become:
f = 1000*e^(Bx)

We still have to find 'B'. So, we use the second condition:
f(1) = 500

Thus,
f = 500 = 1000*e^(B*1)
500 = 1000*(e^B)
e^B = 500/1000 = 1/2
B = ln(1/2)...natural logarithm
B = ln(0.5)

Now, we have found A & B. So, we can plug everything back into our main equation:

f = 1000 * e^(ln(0.5) * x)

The above is the answer you need. But it can also be further simplified:

f = 1000 * { e^(ln(0.5)) }^x
f = 1000 * { 0.5 }^x

f(x) = 1000 * (0.5 ^ x)

Answer 2

It depends on how you define "an exponential function". If we take the function to be in the form f(x) = a*b^(x), and f(0) = 1000 then a*b^0 = 1000, so a=1000. Now set f(1) = 1000*b^(1) = 500, and you get b=1/2. So one answer could be f(x) = 1000*((1/2)^x).

However, if you allow f(x) to take the form of something like ab^(2x), or a(e^x) where e is the exponential number (2.71828...), both of which still look like "exponential functions" to me, you'll get different values for b.

Answer 3

an = a1 * r^x

a0 = 1000 * r^0
1000 = 1000 * 1
1000 = 1000

a1 = 1000 * r^1
500 = 1000 * r
r = (500/1000)
r = (1/2)

f(x) = 1000(1/2)^x
f(x) = 1000/(2^x)
f(x) = 1000 * 2^(-x)
f(x) = 125(2^3) * 2^(-x)
f(x) = 125 * 2^(3 - x)

ANS : f(x) = 125 * 2^(-x + 3)

Answer 4

f(x) = 1000 * (0.5) ^ x

It works.....