# Is this right? e multipled out oddly

1. Nov 18, 2005

### Pengwuino

$$\frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm 2e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }}$$
Is that true???? What am i not realizing here...

Last edited: Nov 18, 2005
2. Nov 18, 2005

### fourier jr

you left out the factor of 2 in the numerator

3. Nov 18, 2005

### Pengwuino

oops yah i did... ill fix that

but is that true? I can't seem to grasp it... and its midnight so i dont think i will be able to easily realize whats going on

I was thinking that it should just be 3e^2t.... and I don't understand how it just got turned into e^4t

4. Nov 18, 2005

### uart

No, $$e^{2t} + 2 e^{2t}$$ would be equal to $$3 e^{2t}$$.
$$e^{2t} * 2 e^{2t} = 2 (e^{2t})^2 = 2 e^{4t}$$

Or for an even more straight forward example,
x + 2x = 3x
x * 2x = 2x^2

Last edited: Nov 18, 2005
5. Nov 18, 2005

### hypermonkey2

youre lucky though, since the $$e^x$$ function was proven to be exponential, thus it follows the rules of exponents, just like anything else. the rest is just algebra.
$$2^2 * 2^2=4 * 4=16=2^4$$
or more generally
$$x^a * x^b=x^{ab}$$ for all bases that correspond to exponential functions. since the "e" function is exponential, you can use these laws.
have fun

Last edited: Nov 18, 2005
6. Nov 18, 2005

### gerben

there is a square missing:
$$e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}$$

7. Nov 18, 2005

### shmoe

$$e^{2t}* e^{2t}=e^{2t+2t}$$, not $$(e^{2t})^{2t}$$

8. Nov 18, 2005

### fourier jr

lol i thought that's what the problem was, where did the 2 go? :tongue2: