Question 694401
This is just an algebra problem involving exponential and logarithmic functions.
It is not as scary as it seems.
You have to ignore the irrelevant stuff added to show you that even higher level math has practical applications for engineers and scientists.
You also have to deal with the way scientist name their variables, but you'll need to deal with that if you take physics.
 
There is calculus mentioned, and physics implied, but most likely those are not your concerns.
{{{T(t)=A+(T[0]-A)e^(-kt)}}} <--> {{{T(t)-A=(T[0]-A)e^(-kt)}}} probably just means that the difference between the temperature of the object, {{{T}}}, and the temperature of the ambient around that object, {{{A}}}, decays exponentially with time.
The little numbers at the bottom (subscripts) are just a way to show that is the same property measured at different times. It's sort of like naming variables with a last name and first name, to have more names available, and to show how some of those variables are related. For scientists, it makes it easier to remember what each variable means. They know that all the T's mean temperature (all relatives, members of the same family) instead of having to remember what was meant by x, y, z, and so on.
 
You just have the exponential function {{{T(t)=A+(T[0]-A)e^(-kt)}}}, and have to deal with algebra and logarithms,while remembering that the name of a variable may include a subscript.
 
For {{{t=0}}} we have
{{{T(0)=A+(T[0]-A)e^(-k*0)=A+(T[0]-A)e^0=A+(T[0]-A)*1=A+T[0]-A=T[0]}}}
For {{{t=t[1]}}} we have
{{{T[1]=T(t[1])=A+(T[0]-A)e^(-kt[1])}}} or simply {{{T[1]=A+(T[0]-A)e^(-kt[1])}}}
Now, back to algebra
{{{T[1]=A+(T[0]-A)e^(-kt[1])}}} --> {{{T[1]-A=(T[0]-A)e^(-kt[1])}}} --> {{{(T[1]-A)/(T[0]-A)=e^(-kt[1])}}}
Time to apply logarithms (base {{{e}}}, of course, natural logarithms, naturally):
{{{(T[1]-A)/(T[0]-A)=e^(-kt[1])}}} --> {{{ln((T[1]-A)/(T[0]-A))=-kt[1]}}}
And now back to simpler algebra, remembering that the team wants to find that constant {{{k}}}:
{{{ln((T[1]-A)/(T[0]-A))=-kt[1]}}} --> {{{(-1/t[1])ln((T[1]-A)/(T[0]-A))=(-1/t[1])(-kt[1])}}} --> {{{highlight(k=(-1/t[1])ln((T[1]-A)/(T[0]-A)))}}} <-->  {{{highlight(k=(1/t[1])ln((T[0]-A)/(T[1]-A)))}}}
That is not the formula you had, so there is a mistake somewhere. Typo? It could be my mistake. It is way past my bedtime, and it's been a rough week.
 
For {{{t=t[n]}}} we have
{{{T[n]=T(t[n])=A+(T[0]-A)e^(-k*t[n])}}} --> {{{T[n]-A=(T[0]-A)e^(-k*t[n])}}} --> {{{(T(n)-A)/(T[0]-A)=e^(-k*t[n])}}}
Taking logarithms:
{{{ln((T[n]-A)/(T[0]-A))=-k*t[n]}}} --> {{{t[n]=-(1/k)ln((T[n]-A)/(T[0]-A))}}}
Again, that is not the formula you had, so there is a mistake somewhere. Typo? It could be my mistake. It is way past my bedtime, and it's been a rough week.