# Help with exponential equation and finding t?

1. Aug 14, 2015

### Dy19

1. The problem statement, all variables and given/known data
The question is: Mr Snail learns 1 fact/day. When will he know as much as the hare?

2. Relevant equations
The hare's learning rate is given by the equation: Learning Rate=50(e^((15-t)/5))+100
Where t is the time in days.
3. The attempt at a solution
Basically I have tried to make the equation equal to t, so that the total facts known will be the same as the amount of days passed, which is what I need. But I'm stuck?
What I have so far is:
t=50(e^((15-t)/5))+100
t-100=50(e^((15-t)/5))
(t/50)-2=e^((15-t)/5)
ln((t/50)-2)=ln(e^((15-t)/5))
ln((t/50)-2)=(15-t)/5
ln((t/50)-2)=3-(t/5)

And I have no idea of how to solve the equation from there?

2. Aug 14, 2015

### Qwertywerty

You need to integrate to get the amount the hare learns in time t .

3. Aug 14, 2015

### Dy19

I have already integrated the equation to get the one I showed?
If a value for t is put into that equation, it will give the amount of facts learnt.

4. Aug 14, 2015

### Qwertywerty

No , you have equated t to learning rate of hare .

5. Aug 14, 2015

### Dy19

I meant it as in that is the total facts learnt. Because the equation has already been integrated.

6. Aug 14, 2015

### Qwertywerty

If you put a value for t in the equation for learing rate of hare , you will get exactly that - the learning rate at time t .

7. Aug 14, 2015

### Qwertywerty

No , that isn't possible . I don't understand this .

8. Aug 14, 2015

### Dy19

Basically, sorry, I must've worded it wrong. But the equation given is for the total amount of facts that the hare learns.

I need help with where I have gotten up to, or if it is wrong (like the way I've gone), I need help as to what to do.

9. Aug 14, 2015

### Ray Vickson

Just to clear things up once and for all: your function $H(t) = 100 + 50 e^{(15-t)/5}$ is NOT the learning rate; it is the (cumulative) amount of Mr. Hare's learning up to time $t$. The learning rate would be $R_H(t) = dH(t)/dt = -10 e^{(15-t)/5}$.

You can see right off that there must be something wrong with the formulation: your $H(t)$ is a decreasing function, so the Hare forgets his knowledge as he ages. On the other hand, if you are really serious that your $H(t)$ is actually the rate of learning, then the Hare's learning up to time $t$ is $\int_0^t H(x) \, dx$, while the Tortoise's learning up to $t$ is just $t$ itself. That means that there is never any time at which the Tortoise catches the Hare.

Anyway, if all you want to do is solve the equation $100 + 50 e^{(15-t)/5} = t$ then just plot both sides to see where the two graphs cross. That will give you an estimate of the root. You can then refine it using one of many numerical equation-solving methods. The equation has no "closed-form" (algebraic formula) solution, although it can be solved in terms of the so-called Lambert W-function; see, eg., https://en.wikipedia.org/wiki/Lambert_W_function . Basically, though, equations such as this are usually solved numerically.