Not that Easy

Algebra Level 5

( 1 + 1 x ) x + 1 = ( 1 + 1 2018 ) 2018 \large \left(1 + \dfrac{1}{x}\right)^{x + 1} = \left(1 + \dfrac{1}{2018}\right)^{2018}

Let α \alpha be the number of real solution(s) to the above equation and β \beta be the sum of all real solution(s). Find α + β \alpha + \beta .

If you think it has no real solution enter 2018.


The answer is -2018.

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1 solution

Chew-Seong Cheong
Aug 25, 2018

( 1 + 1 x ) x + 1 = ( 1 + 1 2018 ) 2018 ( x + 1 x ) x + 1 = ( 2019 2018 ) 2018 = ( 2018 2019 ) 2018 = ( 2019 + 1 2019 ) 2019 + 1 x = 2019 \begin{aligned} \left(1+\frac 1x \right)^{x+1} & = \left(1+\frac 1{2018} \right)^{2018} \\ \left(\frac {x+1}x \right)^{x+1} & = \left(\frac {2019}{2018} \right)^{2018} \\ & = \left(\frac {2018}{2019} \right)^{-2018} \\ & = \left(\frac {-2019+1}{-2019} \right)^{-2019+1} \\ \implies x & = - 2019 \end{aligned}

Therefore, α + β = 1 + ( 2019 ) = 2018 \alpha + \beta = 1 + (-2019) = \boxed{-2018} .

Magnificent!!!

Carlos Baião - 2 years, 3 months ago

How do you know this is the only solution?

Sarthak Sahoo - 1 year, 6 months ago

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The LHS is an increasing function and the RHS is a constant.

Chew-Seong Cheong - 1 year, 6 months ago

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