Question

Suppose f is a function of the form f(x)=aᵇˣ for some constants a and b. The doubling time for f is 1, i.e. f(x + 1) = 2f(x) for all x .
Then f(x) = ____________

Answer 1

f(0) = a^0 = 1
f(1) = a^b = 2
so b = loga(2)
f(x) = a^loga(2) x = 2^x

so f(x) = 2^x
clearly, f(x+1) = 2^(x+1) = 2f(x)

Answer 2

f(x) = aᵇˣ = 2f(x-1) = 2aᵇˣ⁻¹

Since the doubling time for f is 1, we know that f(x) = 2f(x-1), which means that aᵇˣ = 2aᵇˣ⁻¹. Dividing both sides by aᵇˣ⁻¹, we get:

1 = 2

This is not possible, so the given function f(x) = aᵇˣ does not have a doubling time of 1.

Answer 3

You are correct. I apologize for the oversight.

Given f(0) = a^0 = 1 and f(1) = a^b = 2, we have b = logₐ(2). Therefore, f(x) = a^(logₐ(2)*x) = 2^x.

So, f(x) = 2^x satisfies the condition f(x + 1) = 2f(x) for all x.

Thank you for pointing out the correct solution.