What is expand (x+e)x^{e-1}e^{x-e} ?

The solution to expand (x+e)x^{e-1}e^{x-e} is e^{x-e}x^e+e^{1+x-e}x^{e-1}

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expand (x+e)x^{e-1}e^{x-e}

expand

(x + e) x^{e - 1} e^{x - e}

e^{x - e} x^{e} + e^{1 + x - e} x^{e - 1}

Solution steps

(x + e) x^{e - 1} e^{x - e}

= e^{x - e} x^{e - 1} (x + e)

Apply the distributive law:

a (b + c) = ab + a c

a = x^{e - 1} e^{x - e}, b = x, c = e

= x^{e - 1} e^{x - e} x + x^{e - 1} e^{x - e} e

= e^{x - e} x x^{e - 1} + e e^{x - e} x^{e - 1}

Simplify

e^{x - e} x x^{e - 1} + e e^{x - e} x^{e - 1} : e^{x - e} x^{e} + e^{1 + x - e} x^{e - 1}

= e^{x - e} x^{e} + e^{1 + x - e} x^{e - 1}

i x (i x - 1) (i x - 5)

(x - 1)^{(x - 1)^{2}}

\frac{27 ( 8 + π )}{2}

\frac{27 ( 8 + π )}{4}

\frac{1}{( s \cdot ( s ^{2} + 5 ) )}

What is expand (x+e)x^{e-1}e^{x-e} ?
The solution to expand (x+e)x^{e-1}e^{x-e} is e^{x-e}x^e+e^{1+x-e}x^{e-1}