If f is an exponential function, and f(0) = 500, and f(1) = 250, find the defining equation for f.?

Answer 1

f(x) = a . e^bx

500 = a . e^0 = a
250 = 500 . e^b
250/500 = e^b
b = ln(1/2) = -ln(2)

f(x) = 500 e^(-ln(2)x)

Answer 2

f(x) = k* a^x f(0) = k= 500 f(1)= 500*a= 250 so a=1/2
f(x) = 500*(1/2)^x

Answer 3

Let us assume the given exponential function is:
f(x)=ae^(bx)
Given that, f(0)=500
Therefore, a=500.
Also, f(1)=250
i.e. ae^(b)=250 and 500e^(b)=250
i.e. e^(b)=250/500=1/2
b=ln[2^(-1)]=-ln2

Therefore f(x)=500e^(-xln2).

{ OR f(x)=500e^[ln2^(-x)]=500[2^(-x)] = 500/(2^x) }

Answer 4

If f(0) = 500 we know that the form of the equation will be f(x) = 500e^Cx where C is a constant.

Since f(1) = 250, e^C*1 = 0.5...solve for C

Cln(e) = ln(0.5) ---> C = -0.693

So f(x) = 500e^(-0.693x)

Answer 5

The question is unclear.
There can be infinite types of
exponential equations.
Nevertheless let us assume
f[x]=a^x+b
f[0]=1+b=500
b=499
f[1]=a+b=250
a=250-499
=-249
f[x]=-249^x+499

Answer 6

Maybe f(x) = 500/(2^x)