Math Help - Simplify expression

1. Simplify expression

({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 ..

2. Originally Posted by daemonlies
({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 ..
$3^{2x}e^{3x} - 3^{-2x}e^{-3x}$

Show us your work and we can tell you where you went wrong.

-Dan

3. Originally Posted by topsquark
The answer is: . $3^{2x}e^{3x} - 3^{-2x}e^{-3x}$

Dan is correct . . . but how far should we simplify?

$\text{We have: }\; 9^xe^{3x} -\frac{1}{9^xe^{3x}} \;=\; \dfrac{9^{2x}e^{6x} - 1}{9^xe^{3x}} \;=\;\dfrac{81^xe^{6x} - 1}{9^xe^{3x}}$

4. Hi Dan,

Isn't the middle term of the posted function equal to 1 ?

bjh

5. Originally Posted by bjhopper
Hi Dan,

Isn't the middle term of the posted function equal to 1 ?

bjh
Sorry. I'm not sure what you mean by "middle term." Here's the solution:
$((3e)^{2x} - e^{-x})(e^x + (3e)^{-2x})$

$= (3e)^{2x}e^x + (3e)^{2x}(3e)^{-2x} - e^{-x}e^x - e^{-x}(3e)^{-2x}$

$= 3^{2x}e^{2x}e^x + 1 - 1 - 3^{-2x}e^{-x}e^{-2x}$

$= 3^{2x}e^{3x} - 3^{-2x}e^{-3x}$

Please let me know which step you are referring to and we can dissect it.

-Dan

6. Hi Dan,
Thanks for your reply.I did not remove brackets properly and was getting one term in the middle of the other two =1 when I should gotten two terms +1 and -1.

bjh

7. This is what i got .....

(3e^2x - e^(-x) )(e^x+3e^(-2x) )

3e^3x + 9e - e - 3e^(-3x)

Cancelled out 3e^3x & - 3e^(-3x)

9e - e = 8e

8. Originally Posted by daemonlies
This is what i got .....

(3e^2x - e^(-x) )(e^x+3e^(-2x) )

3e^3x + 9e - e - 3e^(-3x)

Cancelled out 3e^3x & - 3e^(-3x)

9e - e = 8e
The original problem as written was stated in terms of $(3e)^{2x}$. In this last post you wrote $3e^{2x}$. Which was it supposed to be?

Otherwise recall the general rule
$(a^n)(a^m) = a^{n + m}$

$e^xe^{-x} = e^{x + -x} = e^0$

by the rule I just mentioned. e^0 = 1, not e.

Similarly
$(3e)^{2x}(3e)^{-2x} = (3e)^{2x + -2x} = (3e)^0 = 1$

Finally
$3e^{3x} - 3e^{-3x} \neq 0$

I'd recommend you review your exponent rules.

-Dan

Edit: If your problem really is
$(3e^{2x} - e^{-x})(e^x + 3e^{-2x})$

$(3e^{2x} - e^{-x})(e^x + 3e^{-2x}) = 3e^{3x} - e^{-3x} + 8$