Is this right? e multipled out oddly...

Pengwuino

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

fourier jr

you left out the factor of 2 in the numerator

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

uart

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

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

hypermonkey2

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

gerben

there is a square missing:
[tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}[/tex]

shmoe

[tex]e^{2t}* e^{2t}=e^{2t+2t}[/tex], not [tex](e^{2t})^{2t}[/tex]

fourier jr

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