Why does $Ce^x = e^{x + \ln(C)}$?

Question

This is probably very simple but I can't figure out why $Ce^x = e^{x + \ln C}$ for some constant $C$?

Answer 1

$$e^{x+\ln C}=e^xe^{lnC}=e^xC$$

$C$ has to be positive for $\ln C$ to make sense.

Answer 2

$$e^{x+\ln(C)}=e^x\cdot e^{\ln(C)}=Ce^x$$

provided $C>0$.

Answer 3

It's wrong for $C\leq0$, but for $C>0$ we have $$Ce^x=e^{\ln{}C}e^x=e^{x+\ln{C}}$$

Answer 4

$Ce^x = e^{\ln Ce^x} = e^{\ln C +\ln e^x} = e^{\ln C+ x}$, since the e-function and the ln-function are inverse to each other on the domain ${\mathbb R}_{>0}$.

Answer 5

(Assuming $C>0$): Let $f(x)=Ce^x$. Then $\ln f(x)=\ln (Ce^x)=\ln C+\ln e^x=\ln C +x$. Then $f(x)=e^{\ln C+x}$.