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

by Pengwuino
Tags: multipled
 PF Gold P: 7,120 $$\frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm 2e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }}$$ Is that true???? What am i not realizing here...
P: 948
 Quote by Pengwuino $$\frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }}$$ Is that true???? What am i not realizing here...
you left out the factor of 2 in the numerator
 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
P: 2,751
Is this right? e multipled out oddly...

 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, $$e^{2t} + 2 e^{2t}$$ 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
 P: 102 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
 P: 540 there is a square missing: $$e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}$$
 Quote by gerben there is a square missing: $$e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}$$
$$e^{2t}* e^{2t}=e^{2t+2t}$$, not $$(e^{2t})^{2t}$$