Exponential modelling with given points

Question

Given the points (2,3) and (5,54) are on an exponential model, find the equation for this model in the form $f(x) = C\cdot e^{kx}$

So by plugging in the values given, I found that $k=\frac{\ln(18)}3$ and that $C = \frac{3}{e^{\frac23\cdot \ln(18)}}$, but the resulting equation does not yield the given points... How can I solve this problem?

Answer 1

Solving results in $k = 1/3 \ln 18$ and $C = 3(18)^{-2/3}$. Substituting gives \begin{align} f(2) & = 3(18)^{-2/3}e^{2/3 \ln 18} = 3 \\ f(5) & = 3(18)^{-2/3}e^{5/3 \ln 18} = 3(18) = 54, \end{align} your answer seems to be correct.

Answer 2

but the resulting equation does not yield the given points...

I fixed your notation and then you get $$ f(x)=\frac3{e^{\frac23\ln(18)}}e^{\frac{\ln(18)}3x}=3e^{-\frac23\ln(18)}e^{\frac{\ln(18)}3x}=3e^{\frac13\ln(18)(x-2)}. $$ Now it is maybe easier to check $$ f(2)=3e^0=3\text{ and }f(5)=3e^{\frac13\ln(18)3}=3e^{\ln(18)}=3\cdot 18=54. $$