What is (e^{9999}+1)^{-2/9999} ?

The solution to (e^{9999}+1)^{-2/9999} is 1/((e^{9999)+1)^{2/9999}}

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simplify (e^{9999}+1)^{-2/9999}

simplify

(e^{9999} + 1)^{- \frac{2}{9999}}

\frac{1}{( e ^{9999} + 1 ) ^{\frac{2}{9999}}}

Solution steps

(e^{9999} + 1)^{- \frac{2}{9999}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{( e ^{9999} + 1 ) ^{\frac{2}{9999}}}

\frac{- 190}{10}

(1.1 \cdot 1 0^{- 3}) \cdot 9.8

1.33 \cdot 1 0^{- 6}

\frac{n ^{3} + 4}{( n + 1 ) ^{3} + 4}

What is (e^{9999}+1)^{-2/9999} ?
The solution to (e^{9999}+1)^{-2/9999} is 1/((e^{9999)+1)^{2/9999}}