# Continuous compunding

If I start with Rs 100 and it becomes Rs 200 in a year, why is the continuous compounding rate obtained by doing Ln(200/100)? Could someone please help me?

V_t = V_0e^{r_{cont}t}
e^{r_{cont}t} = \frac{V_t}{V_0}
r_{cont}t = ln\left(\frac{V_t}{V_0}\right)
r_{cont} = \frac1tln\left(\frac{V_t}{V_0}\right)
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I can’t understand this. Could you please simplify it?

It can’t be simplified.

I can explain it.

V_0 is the value at time t = 0; in your case, that’s Rs 100.

V_t is the value at some time t in the future; in your case, that’s Rs 200 at time t = 1.

The first equation says that the value V_t at any time in the future is the value at time 0 increased at the continuous compounding rate r_{cont} for the amount of time t. We know t = 1, V_0 = 100, and V_1 = 200; we’re trying to solve for r_{cont}.

First, we divide each side by V_0.

Then, we take the natural logarithm of each side, remembering that ln\left(e^x\right) = x.

Finally, we divide each side by t.

Using the values we know,

r_{cont} = \frac1tln\left(\frac{V_t}{V_0}\right) = \frac11ln\left(\frac{200}{100}\right) = ln\ 2 = 0.693147 = 69.3147\%

Thank you!

My pleasure.