# simplify exponential problem...

• Oct 10th 2010, 06:12 AM
wolfhound
simplify exponential problem...
Hi, I need some help step by step with simplifying this please...

$e^x^2 - 4(e^x)^2 + (-3)e^x^2 + (2ee^2^x^-^1)$
• Oct 10th 2010, 06:49 AM
skeeter
Quote:

Originally Posted by wolfhound
Hi, I need some help step by step with simplifying this please...

$e^x^2 - 4(e^x)^2 + (-3)e^x^2 + (2ee^2^x^-^1)$

if this is the expression you meant to post ...

$e^{x^2} - 4(e^x)^2 + (-3)e^{x^2} + (2e \cdot e^{2x-1})=$

$e^{x^2} - 4e^{2x} - 3e^{x^2} + 2e^{2x}$

finish by combining like terms.
• Oct 10th 2010, 08:16 AM
wolfhound
Quote:

Originally Posted by skeeter
if this is the expression you meant to post ...

$e^{x^2} - 4(e^x)^2 + (-3)e^{x^2} + (2e \cdot e^{2x-1})=$

$e^{x^2} - 4e^{2x} - 3e^{x^2} + 2e^{2x}$

finish by combining like terms.

Would the final answer be this $-2e^x^2 -2e^2^x$?
So the -1 exponent disappears at the end because 2e^+1 is multiplying it out?

Thanks
• Oct 10th 2010, 08:23 AM
Unknown008
You can further factorise out -2, to give:

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

$2ee^{2x-1} = 2e.e^{2x}.e^{-1} = \dfrac{2e.e^{2x}}{e}$

e divided by e is 1.

Or you go with combining the powers:

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