(1 point) Suppose \( f \) is a function of the form \[ f(x)=a^{b x} \] for some constants \( a \) and \( b \). The doubling time for \( f \) is 3 , i.e., \[ f(x+3)=2 f(x) \] for all \( x \) Then \( f(x)=\square \)

To solve for \( f(x) \) given that the doubling time is 3, we start from the function \( f(x) = a^{b x} \). According to the doubling time, we can write: \[ f(x+3) = a^{b(x+3)} = a^{bx + 3b} = a^{bx} \cdot a^{3b} \] Setting this equal to \( 2 f(x) \) gives us: \[ a^{bx} \cdot a^{3b} = 2 a^{bx} \] We can divide both sides by \( a^{bx} \) (assuming \( a^{bx} \neq 0 \)), resulting in: \[ a^{3b} = 2 \] Taking the logarithm, we find: \[ 3b \log(a) = \log(2) \] \[ b \log(a) = \frac{\log(2)}{3} \] Now, choosing any positive value for \( a \) allows us to express \( b \). When \( a = 2^{1/3} \) (the cube root of 2), then: \[ b \log(2^{1/3}) = \frac{\log(2)}{3} \] \[ b \cdot \frac{1}{3} \log(2) = \frac{\log(2)}{3} \] \[ b = 1 \] Thus, substituting \( a \) and \( b \) into the equation gives: \[ f(x) = \left(2^{1/3}\right)^{1 \cdot x} = 2^{x/3} \] So we arrive at: \[ f(x) = 2^{x/3} \]