201 9 n 2019 ^n

Algebra Level 3

2019 2019 2019 2019 2019 2019 2019 2019 = 201 9 n \large \sqrt[2019]{{2019} \sqrt[2019]{{2019} \sqrt[2019]{{2019}\sqrt[2019]{{2019} \cdots } }}}= 2019^n

If the equation above holds true for some real n n , what is the value of 1 + 1 n 1 + \dfrac 1n ?


The answer is 2019.

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1 solution

Hugh Sir
Dec 28, 2018

2019 2019 2019 2019 2019 2019 2019 2019 = 201 9 n \large \sqrt[2019]{{2019} \sqrt[2019]{{2019} \sqrt[2019]{{2019} \sqrt[2019]{{2019} \cdots }}}} = 2019^{n}

2019 × 201 9 n 2019 = 201 9 n \large \sqrt[2019]{{2019} \times 2019^{n}} = 2019^{n}

201 9 n + 1 2019 = 201 9 n \large 2019^{\frac{n+1}{2019}} = 2019^{n}

n + 1 2019 = n \dfrac{n+1}{2019} = \large n

n + 1 n = 2019 \dfrac{n+1}{n} = 2019

Therefore, 1 + 1 n = 2019 1 + \dfrac{1}{n} = 2019 .

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