Solve-for-real-numbers-2-x-2-1-1-7-1-2-x-2-1-1-7-

Question Number 161884 by HongKing last updated on 23/Dec/21

Solve for real numbers :

\sqrt[7]{2^{x} + 2^{- 1}} = 1 + \sqrt[7]{2^{x} - 2^{- 1}}

Commented by mr W last updated on 23/Dec/21

\sqrt[7]{2^{x} + \frac{1}{2}} = 1 + \sqrt[7]{2^{x} - \frac{1}{2}}

1 = 1 + 0 ?

\Rightarrow 2^{x} = \frac{1}{2} ✓

\Rightarrow x = - 1

Commented by HongKing last updated on 25/Dec/21

thank you my dear Sir