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

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$2^{\mathbf{x}}+2^{1+\frac{1}{\sqrt{\mathbf{x}}}}=6$$$$$$

Commented by HongKing last updated on 07/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Ser}\mathrm{cool}$$

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

$$6=2^1+2^{2}or6=2^2+2^1$$$$x=1\wedge 1+\frac{1}{\sqrt{x}}=2\left(Satisfied\right)$$$$x=2\wedge 1+\frac{1}{\sqrt{x}}=1\left(Contradictory\right)$$$$x=1$$

Answered by ghimisi last updated on 07/Nov/21

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$2^{\mathbf{x}}+2^{1+\frac{1}{\sqrt{\mathbf{x}}}}=6$$$$2^x+2\cdot 2^{\frac{1}{\sqrt{x}}}=2^x+2^{\frac{1}{\sqrt{x}}}+2^{\frac{1}{\sqrt{x}}}⩾3\sqrt[3]{2^x\cdot 2^{\frac{1}{\sqrt{x}}}\cdot 2^{\frac{1}{\sqrt{x}}}}=$$$$=3\sqrt[3]{2^{x+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}}}\stackrel{}{⩾}3\sqrt[3]{2^{3\sqrt[3]{x\cdot \frac{1}{\sqrt{x}}\cdot \frac{1}{\sqrt{x}}}}}=6\Rightarrow$$$$x=\frac{1}{\sqrt{x}}\Rightarrow x^3=1\Rightarrow x=1$$

Commented by HongKing last updated on 07/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{vert}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Ser}\mathrm{perfect}$$

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