You're right that you use the natural log function, but your form is incorrect. What would happen if you take the natural log of the exponential function, or ln(e^x) = ?
Let's do it first with the more familiar base 10. log( 10^2 ) = 2 (right?) log( 10^x ) = ? ln( e^x ) is analogous to log( 10^x )...
log(10^x) = x ? analogous?? The same as?? but log e = 1 does that mean log e^x = 1^x ? therefore log e^x = x ?
No, you need to keep your log() and ln() straight. log() is used with base 10 math, and ln() is used with base e math. log( 10^x ) = x ln( e^x ) = ? Try a few numbers on your calculator to help you keep it straight. There's a reason that most calculators overload the log() key with 10^x and overload the ln() key with e^x.....
How come you did not take the natural log on the left hand side? :surprised If this is your original equation, then: [tex]\ln V = \ln V_0 - {t\over RC}[/tex] Furthur evaluation is simple.