Simplifying ln(c*e^{-kt})

[vorcil]

If i were to simplify

$$ln(c*e^{-kt})$$

what happens?

do I get

$$c*(-kt)$$
or
$$ln(c) * (-kt)$$

or something else?

I'm not sure

 

[Dick]

Neither. log(a*b)=log(a)+log(b). Start with that.

 

[vorcil]

Neither. log(a*b)=log(a)+log(b). Start with that.

Cheers big ears,

 

[jegues]

$$ln(AB) = ln(A) + ln(B)$$

Does that help?

EDIT:

Neither. log(a*b)=log(a)+log(b). Start with that.

You beat me to it!