simplify exponential problem...

**wolfhound** · October 10th 2010, 05:12 AM

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

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

**skeeter** · October 10th 2010, 05:49 AM

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.

**wolfhound** · October 10th 2010, 07:16 AM

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

**Unknown008** · October 10th 2010, 07:23 AM

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:

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

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