Menu Close

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:  ((2^x  + 2^(-1) ))^(1/7)  = 1 + ((2^x  - 2^(-1) ))^(1/7)
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt[{\mathrm{7}}]{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{-\mathrm{1}} }\:=\:\mathrm{1}\:+\:\sqrt[{\mathrm{7}}]{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:-\:\mathrm{2}^{-\mathrm{1}} } \\ $$
Commented by mr W last updated on 23/Dec/21
((2^x +(1/2)))^(1/7) =1+((2^x −(1/2)))^(1/7)   1=1+0 ?  ⇒2^x =(1/2) ✓  ⇒x=−1
$$\sqrt[{\mathrm{7}}]{\mathrm{2}^{{x}} +\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{1}+\sqrt[{\mathrm{7}}]{\mathrm{2}^{{x}} −\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{1}=\mathrm{1}+\mathrm{0}\:? \\ $$$$\Rightarrow\mathrm{2}^{{x}} =\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$$$\Rightarrow{x}=−\mathrm{1} \\ $$
Commented by HongKing last updated on 25/Dec/21
thank you my dear Sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *