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

Solve for real numbers: 2^x + 2^(1 + (1/( (√x)))) = 6

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

Commented by HongKing last updated on 07/Nov/21

thank you very much my dear Ser cool

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

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

6=2^1 +2^2 or 6=2^2 +2^1 x=1 ∧ 1+(1/( (√x)))=2 (Satisfied) x=2 ∧ 1+(1/( (√x)))=1 (Contradictory) x=1

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

Answered by ghimisi last updated on 07/Nov/21

Solve for real numbers: 2^x + 2^(1 + (1/( (√x)))) = 6 2^x +2∙2^(1/( (√x))) =2^x +2^(1/( (√x))) +2^(1/( (√x))) ≥3 ((2^x ∙2^(1/( (√x))) ∙2^(1/( (√x))) ))^(1/3) = =3(2^(x+(1/( (√x)))+(1/( (√x)))) )^(1/3) ≥^ 3(2^(3((x∙(1/( (√x)))∙(1/( (√x)))))^(1/3) ) )^(1/3) =6⇒ x=(1/( (√x)))⇒x^3 =1⇒x=1

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}}}} \:=\:\mathrm{6} \\ $$$$\mathrm{2}^{{x}} +\mathrm{2}\centerdot\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} =\mathrm{2}^{{x}} +\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} +\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} \:\geqslant\mathrm{3}\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{{x}} \centerdot\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} \centerdot\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{{x}}}} }= \\ $$$$=\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}^{{x}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}} }\overset{} {\geqslant}\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\centerdot\frac{\mathrm{1}}{\:\sqrt{{x}}}\centerdot\frac{\mathrm{1}}{\:\sqrt{{x}}}}} }=\mathrm{6}\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\:\sqrt{{x}}}\Rightarrow{x}^{\mathrm{3}} =\mathrm{1}\Rightarrow{x}=\mathrm{1} \\ $$

Commented by HongKing last updated on 07/Nov/21

thank you vert much my dear Ser perfect

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

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