How can the log in the denominator of the charging equation be simplified?

[Dr. Ficus]

Homework Statement

Basically, I'm just trying to verify that I can do basic algebraic manipulation correctly. The objective: solve the charging equation of a simple RC circuit for (C), capacitance.

Homework Equations

[tex] v\left(t\right)={v}_{s}\left(1-{e}^{{-t}/{RC}}\right) [/tex]

The Attempt at a Solution

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I ended up with the following.
[tex] C=\frac{-t}{R\mathrm{ln}\left(\frac{{v}_{s}-{v}_{\left(t\right)}}{{v}_{s}}\right)} [/tex]
Is this correct? Also, I have a suspicion that the log in the denominator can somehow be simplified more than just putting [tex] \mathrm{ln}\left(1-\frac{v\left(t\right)}{{v}_{s}}\right) [/tex] but I can't think of how. Thanks in advance!

 

[AssyriaQ]

That is correct. You can, if you so prefer, use the principle
[tex]\ln(\frac{a}{b})=\ln a - \ln b[/tex]
for the log in the denominator.