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Is this right? e multipled out oddly...

by Pengwuino
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Pengwuino
#1
Nov18-05, 01:27 AM
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[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...
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fourier jr
#2
Nov18-05, 01:44 AM
P: 948
Quote Quote by Pengwuino
[tex]
\frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }}
[/tex]
Is that true???? What am i not realizing here...
you left out the factor of 2 in the numerator
Pengwuino
#3
Nov18-05, 01:54 AM
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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
#4
Nov18-05, 08:18 AM
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Is this right? e multipled out oddly...

Quote Quote by 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
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
#5
Nov18-05, 09:51 AM
P: 102
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
#6
Nov18-05, 09:53 AM
P: 540
there is a square missing:
[tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}[/tex]
shmoe
#7
Nov18-05, 10:26 AM
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Quote Quote by gerben
there is a square missing:
[tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}[/tex]
[tex]e^{2t}* e^{2t}=e^{2t+2t}[/tex], not [tex](e^{2t})^{2t}[/tex]
fourier jr
#8
Nov18-05, 10:56 AM
P: 948
Quote Quote by 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
lol i thought that's what the problem was, where did the 2 go?


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