[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...
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
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