Solve-for-real-numbers-2-x-2-1-1-x-6-

Question Number 158644 by HongKing last updated on 07/Nov/21

Solve for real numbers :

2^{x} + 2^{1 + \frac{1}{\sqrt{x}}} = 6

Commented by HongKing last updated on 07/Nov/21

thank you very much my dear Ser cool

Commented by Rasheed.Sindhi last updated on 07/Nov/21

6 = 2^{1} + 2^{2} o r 6 = 2^{2} + 2^{1}

x = 1 \land 1 + \frac{1}{\sqrt{x}} = 2 (S a t i s f i e d)

x = 2 \land 1 + \frac{1}{\sqrt{x}} = 1 (C o n t r a d i c t o r y)

x = 1

Answered by ghimisi last updated on 07/Nov/21

Solve for real numbers :

2^{x} + 2^{1 + \frac{1}{\sqrt{x}}} = 6

2^{x} + 2 \cdot 2^{\frac{1}{\sqrt{x}}} = 2^{x} + 2^{\frac{1}{\sqrt{x}}} + 2^{\frac{1}{\sqrt{x}}} ⩾ 3 \sqrt[3]{2^{x} \cdot 2^{\frac{1}{\sqrt{x}}} \cdot 2^{\frac{1}{\sqrt{x}}}} =

= 3 \sqrt[3]{2^{x + \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}}}} \overset{}{⩾} 3 \sqrt[3]{2^{3 \sqrt[3]{x \cdot \frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}}}} = 6 \Rightarrow

x = \frac{1}{\sqrt{x}} \Rightarrow x^{3} = 1 \Rightarrow x = 1

Commented by HongKing last updated on 07/Nov/21

thank you vert much my dear Ser perfect