The problem is:
$\displaystyle e^{lnx} = 4$
Is there a way I can rewrite this to solve for x, or some rule that I am suppose to apply?
@Plato:
Well, clearly it is not a simple problem for me, especially seeing that I have not dealt with a problem like this for two semester. And if it is such a waste, why spend the time replying?
@Skeeter:
Now I believe I remember seeing something like this. And isn't true that the base e and the exponent go away because they are inverses? That's the part I have a little trouble wrapping my head around. That's the only explanation I have been given, and it just seem a good enough explanation for me.
What I started to do was: $\displaystyle ln(e)^{lnx} = ln(4)\rightarrow\ln(x)\times\ln(e) = ln(4)\rightarrow\ln(x) = ln(4)$, then finally $\displaystyle e^{ln(4)} = x$
which sort of brought me back to square one. I know I can just evaluate it in the calculator, but I don't know why the answer comes out the way it does.
BashyBoy, I think that Plato was a bit annoyed that you would post a question like that, involving exponential and logarithms, without finding out, or reviewing, what exponentials and logarithms are. And the whole point of the problem is the definition of 'exponential' and 'logarithm'.