What is (4e^{x^2+1})(4e^{x^2}) ?

The solution to (4e^{x^2+1})(4e^{x^2}) is 16e^{2x^2+1}

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simplify (4e^{x^2+1})(4e^{x^2})

simplify

(4 e^{x^{2} + 1}) (4 e^{x^{2}})

16 e^{2 x^{2} + 1}

Solution steps

(4 e^{x^{2} + 1}) (4 e^{x^{2}})

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

4 \cdot 4 = 4^{1 + 1}

= e^{x^{2} + 1} \cdot 4^{1 + 1} e^{x^{2}}

Add the numbers:

1 + 1 = 2

= e^{x^{2} + 1} \cdot 4^{2} e^{x^{2}}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{x^{2} + 1} e^{x^{2}} = e^{x^{2} + 1 + x^{2}}

= 4^{2} e^{x^{2} + 1 + x^{2}}

Simplify

x^{2} + 1 + x^{2} : 2 x^{2} + 1

= 4^{2} e^{2 x^{2} + 1}

4^{2} = 16

= 16 e^{2 x^{2} + 1}

1.9 \cdot 1 0^{- 5} (3)

(2 x + 3)^{2} (4 x - 2)^{2}

(\frac{1}{s - 4 ^{e}})^{- 2 s}

What is (4e^{x^2+1})(4e^{x^2}) ?
The solution to (4e^{x^2+1})(4e^{x^2}) is 16e^{2x^2+1}