({3e}^{2x} - {e}^{-x})({e}^{x} + {3e}^{-2x})
Need help with the above equation ... Remove brackets and simplify expontential expression !!!
I got a result of .... after simplifying .... 8e ... i hope its correct ..
Sorry. I'm not sure what you mean by "middle term." Here's the solution:
$\displaystyle ((3e)^{2x} - e^{-x})(e^x + (3e)^{-2x})$
$\displaystyle = (3e)^{2x}e^x + (3e)^{2x}(3e)^{-2x} - e^{-x}e^x - e^{-x}(3e)^{-2x}$
$\displaystyle = 3^{2x}e^{2x}e^x + 1 - 1 - 3^{-2x}e^{-x}e^{-2x}$
$\displaystyle = 3^{2x}e^{3x} - 3^{-2x}e^{-3x}$
Please let me know which step you are referring to and we can dissect it.
-Dan
The original problem as written was stated in terms of $\displaystyle (3e)^{2x}$. In this last post you wrote $\displaystyle 3e^{2x}$. Which was it supposed to be?
Otherwise recall the general rule
$\displaystyle (a^n)(a^m) = a^{n + m}$
Let's first talk about the
$\displaystyle e^xe^{-x} = e^{x + -x} = e^0$
by the rule I just mentioned. e^0 = 1, not e.
Similarly
$\displaystyle (3e)^{2x}(3e)^{-2x} = (3e)^{2x + -2x} = (3e)^0 = 1$
Finally
$\displaystyle 3e^{3x} - 3e^{-3x} \neq 0$
I'd recommend you review your exponent rules.
-Dan
Edit: If your problem really is
$\displaystyle (3e^{2x} - e^{-x})(e^x + 3e^{-2x})$
then the answer is
$\displaystyle (3e^{2x} - e^{-x})(e^x + 3e^{-2x}) = 3e^{3x} - e^{-3x} + 8$
See if you can get that with what we showed you in this thread.