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This is probably very simple but I can't figure out why $Ce^x = e^{x + \ln C}$ for some constant $C$?
5 Answers
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$$e^{x+\ln C}=e^xe^{lnC}=e^xC$$
$C$ has to be positive for $\ln C$ to make sense.
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$$e^{x+\ln(C)}=e^x\cdot e^{\ln(C)}=Ce^x$$
provided $C>0$.
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It's wrong for $C\leq0$, but for $C>0$ we have $$Ce^x=e^{\ln{}C}e^x=e^{x+\ln{C}}$$
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$Ce^x = e^{\ln Ce^x} = e^{\ln C +\ln e^x} = e^{\ln C+ x}$, since the e-function and the ln-function are inverse to each other on the domain ${\mathbb R}_{>0}$.
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(Assuming $C>0$): Let $f(x)=Ce^x$. Then $\ln f(x)=\ln (Ce^x)=\ln C+\ln e^x=\ln C +x$. Then $f(x)=e^{\ln C+x}$.
