Differential equations

They work out the rate constant
k = 1/5 ln(2)
and the exponential is
e^(-kt)
so the -kt exponent is
-kt = -t/5 ln(2) = ln(2^(-t/5)) = ln( (1/2)^(t/5))
Then sub into the differentia equation for V and the e and ln are inverse functions.

Thank you. I understand that but would not have been able to work it out for myself.

I would like to rate you but as you are one of the few people to reply I have to find other before I can press the + again.

NP. Couple of things. Firstly, its unnecessary algebra. You could simply put
V = 20/k - 20/k e^(-kt)
with the value of k youve worked out and hit the calculator.

Secondly, you can always write an exponentiall
e^(kt) = a^t
for some base a. Here k can be positive or negative. The scalar k is just scaling time, just as the value of the base a does. Youo can always write a in terms of k or vice versa which is "simply" all theyve done here and a=(1/2)^(1/5), so the fifth root of 1/2 which is a bit less than 1. So its a slowly decaying exponential which corresponds to k being negative and close to zero..

For the latter point, having an idea that you can write it in either form is arguably the most important thing, after that its just a bit of tedious algebra to transform k <-> a.

Thirdly, you could note that t=5 is the half life of the process as dv/dt goes from 20->10. So for t=10, this must represent a decay to (1/2)^2 = 1/4 of the initial value, hence the 1/4 multiplier near the bottom. After t=15, it would be 1/8 etc ... Saves a fair bit of heartache.