## capacitance equation

how would i rearrange this equation to find the capacitance, c?

V=Vo exp(-t/RC)

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 Do you know what the inverse of the exponential function is?
 would it go to, V = Vo ln + (-t/RC) then to, V = ln Vo - t x 1/RC

## capacitance equation

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) = ?

 ln(e) + x ????? not really sure you've lost me a bit sorry

Mentor
 Quote by dan greig ln(e) + x ????? not really sure you've lost me a bit sorry
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 ?

Mentor
 Quote by dan greig 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.....

 Quote by dan greig would it go to, V = Vo ln + (-t/RC) then to, V = ln Vo - t x 1/RC
How come you did not take the natural log on the left hand side?
If this is your original equation, then:
 V=Vo exp(-t/RC)
$$\ln V = \ln V_0 - {t\over RC}$$

Furthur evaluation is simple.