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

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

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

expand

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

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

Solution steps

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

Apply the distributive law:

a (b + c) = ab + a c

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

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

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

Simplify

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

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

\frac{( 40 - k ) ( 5 + k )}{45}

(10 x^{4} - 11 x^{2} + 1)^{5}

(x - 3)^{2} + y^{2} + (x + 3)^{2} + y^{2}

(2 + x + x^{4}) (5 x - \frac{1}{2 x ^{3}})^{8}

A (- 4, - 5) B (- 2, 3) C (1, - 2)

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