Can an Analytic Method Solve for t in the Equation 2e^(t²) + e^t - 3e = 0?

[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!

 

[kreil]

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

 

[mishcake]

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

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

 

[rasmhop]

Is your equation:
$$2(e^t)^2 +e^t - 3e = 0$$
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.

 

[mishcake]

Is your equation:
$$2(e^t)^2 +e^t - 3e = 0$$
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.

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

 

[andylu224]

obviously, one solution is t = 1.

 

[nicksauce]

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

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.

 

[kreil]

oops