Question involving exponentials and natural logs?

Question

I am given
V(1) = Vf * [1 - exp(-t1 / T )]
and
V(2) = Vf * [1 - exp(-t2 / T )]

solve for T in terms of V(1), V(2), t1, t2

Answer 1

Vf = V(1) / [1 - exp(-t1 / T )] = V(2) / [1 - exp(-t2 / T )]
V(1) [1 - exp(-t2 / T )] = V(2) [1 - exp(-t1 / T )]
V(1) - V(1)exp(-t2 / T ) = V(2) - V(2)exp(-t1 / T )
V(1) - V(2) = V(1)exp(-t2 / T ) - V(2)exp(-t1 / T )
At this point one realizes that T cannot be isolated algebraically, so a solution for T is transcendental.
- V(1) + V(2) = - V(1)exp(-t2 / T ) + V(2)exp(-t1 / T )
V(2)exp(-t1 / T ) = V(2) - V(1)[1 - exp(-t2 / T )]
exp(-t1 / T ) = 1 - (V(1)/V2)[1 - exp(-t2 / T )]
-t1 / T = ln(1 - (V(1)/V2)[1 - exp(-t2 / T )])
T = - t1/ln(1 - (V(1)/V2)[1 - exp(-t2 / T )])
From here you have start for numerical solution.

Answer 2

You can use basic algebra to turn the first into exp(-t1/T) into something involving V(1) and Vf. you then take natural log of both sides to get -t1/T = etc. Similarly for the second equation. Rearrage your first result to give Vf = etc which you replace in your second result to eliminate Vf. I'm not prepared to just do it for you - but this should help you do it for yourself which, in the end, will be more use to you.

Answer 3

The be conscious 'organic' is probable a misnomer. Base 10 is quite much less puzzling to apply for undemanding calculations, yet e comes into arithmetic very quickly and for many purposes e is the backside that fits the mathematicians. it quite is all to do with the exponential function and all its ramifications. seek for suggestion from a textual content fabric e book in this and in simple terms slightly math background. Napier did not use e as his base while he invented logarithms. Neither did he use base 10.