# Something I should really know by now

1. Nov 1, 2005

### joex444

Idk if this is right
$$e^{n^2} = (e^n)^n$$
Atleast I didn't think $$e^{n^2} = e^n*e^n$$

I had a test in calc 3 today, and that was part of a series. If it is equal (in the first equation) then I did it right, otherwise not. I came up with convergent, and someone else said they used an integral and got convergent, so...my answer hinges on that, which i should really know by now :)

Last edited: Nov 1, 2005
2. Nov 1, 2005

### hypermorphism

In general, $$a^{b^c} \ne a^{bc}$$.
$$(a^b)^c = a^{bc}$$
You should be able to show this yourself. Your first equality is correct.

3. Nov 1, 2005

### moose

Well, when in doubt just substitute some numbers! I don't understand why people don't do this more often when in doubt.

Like pretend e=2 and that n=3

$$2^9=(2^3)^3$$

we all know $$2^9=512$$ and $$2^3=8$$
$$8^3=512$$
so it works

Now, this is aside from the fact that this is a basic algebraic rule...

4. Nov 1, 2005

### joex444

yea, lol, i just typed in $$({e^n})^n$$ into my calculator and it spit back $$e^{n^2}$$... it's a ti-89.

sometimes you just forget the basic stuff...

5. Nov 1, 2005

### HallsofIvy

In that case, it's good to have a calculator that is smarter than you are!