# Possibly impossible question?

1. Jan 1, 2010

### mishcake

Given that 2e^(t²) + e^t - 3e = 0, solve for t.

I've been racking my brains to figure out a solution to this one. I know that this can easily be solved by using a graphing calculator, but is there an analytic method to deduce the value of t as well? According to my friend, since this is neither a polynomial nor an algebraic expression, an analytic method would be impossible (at least at A levels).

Thanks in advance to all those who help!

Last edited: Jan 1, 2010
2. Jan 1, 2010

### kreil

Try taking the natural logarithm of both sides then using the quadratic equation on the resulting second degree polynomial

3. Jan 1, 2010

### mishcake

But that would only work if the first term is to the power 2t instead of t².

4. Jan 1, 2010

### rasmhop

$$2(e^t)^2 +e^t - 3e = 0$$
or
$$2e^{(t^2)} + e^t -3e = 0$$
? I suspect the former as that's a very common type of problem in elementary algebra and in that case you can indeed get by using the substitution y=e^t.

5. Jan 1, 2010

### mishcake

The latter. If it were the former I wouldn't even be posting it here :p

6. Jan 1, 2010

### andylu224

obviously, one solution is t = 1.

7. Jan 1, 2010

### nicksauce

The natural logarithm is 0 is not defined, and the logarithm of a sum is not the sum of the logarithms, so this does not work at all.

8. Jan 2, 2010

oops