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Is this right? e multipled out oddly...by Pengwuino
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#1
Nov1805, 01:27 AM

PF Gold
P: 7,120

[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... 


#2
Nov1805, 01:44 AM

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#3
Nov1805, 01:54 AM

PF Gold
P: 7,120

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
Nov1805, 08:18 AM

Sci Advisor
P: 2,751

Is this right? e multipled out oddly...
[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 


#5
Nov1805, 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 


#6
Nov1805, 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] 


#7
Nov1805, 10:26 AM

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#8
Nov1805, 10:56 AM

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