$2019 ^n$

Algebra Level 3

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

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

1 solution

Hugh Sir
Dec 28, 2018

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

$\large \sqrt[2019]{{2019} \times 2019^{n}} = 2019^{n}$

$\large 2019^{\frac{n+1}{2019}} = 2019^{n}$

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

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

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