2-3-x-2-3-x-4-x-

Question Number 179529 by mathlove last updated on 30/Oct/22

$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}} +\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} =\mathrm{4}\:\:\:\:\:\:{x}=? \\ $$

Answered by mr W last updated on 30/Oct/22

(2−(√3))(2+(√3))=2^2 −((√3))^2 =4−3=1 ⇒2−(√3)=(1/(2+(√3))) (1/((2+(√3))^x ))+(2+(√3))^x =4 let t=(2+(√3))^x (1/t)+t=4 t^2 −4t+1=0 t=2±(√3) (2+(√3))^x =2±(√3)=2+(√3), (1/(2+(√3))) ⇒x=±1

$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)=\mathrm{2}^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{4}−\mathrm{3}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}−\sqrt{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} }+\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} =\mathrm{4} \\ $$$${let}\:{t}=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} \\ $$$$\frac{\mathrm{1}}{{t}}+{t}=\mathrm{4} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} =\mathrm{2}\pm\sqrt{\mathrm{3}}=\mathrm{2}+\sqrt{\mathrm{3}},\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\pm\mathrm{1} \\ $$

Commented by mathlove last updated on 30/Oct/22

$${thanks}\:{mr} \\ $$

Answered by cortano1 last updated on 30/Oct/22

let (2−(√3))^x = y ⇒(2+(√3))^x =(1/y) ⇒y+(1/y) = 4 ⇒y^2 −4y+1=0 ⇒(y−2)^2 = 3 ⇒y=2+(√3) or y=2−(√3) { ((y=2−(√3) ⇒x=1)),((y=2+(√3)⇒x=−1)) :}

$$\:\mathrm{let}\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:=\:\mathrm{y}\:\Rightarrow\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} =\frac{\mathrm{1}}{\mathrm{y}} \\ $$$$\:\Rightarrow\mathrm{y}+\frac{\mathrm{1}}{\mathrm{y}}\:=\:\mathrm{4}\:\Rightarrow\mathrm{y}^{\mathrm{2}} −\mathrm{4y}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} =\:\mathrm{3} \\ $$$$\Rightarrow\mathrm{y}=\mathrm{2}+\sqrt{\mathrm{3}}\:\mathrm{or}\:\mathrm{y}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\begin{cases}{\mathrm{y}=\mathrm{2}−\sqrt{\mathrm{3}}\:\Rightarrow\mathrm{x}=\mathrm{1}}\\{\mathrm{y}=\mathrm{2}+\sqrt{\mathrm{3}}\Rightarrow\mathrm{x}=−\mathrm{1}}\end{cases} \\ $$

Answered by Frix last updated on 30/Oct/22

b^x +b^(−x) =y; b>0∧y≥2 2cosh (xln b) =y x=((cosh^(−1) (y/2))/(ln b))

$${b}^{{x}} +{b}^{−{x}} ={y};\:{b}>\mathrm{0}\wedge{y}\geqslant\mathrm{2} \\ $$$$\mathrm{2cosh}\:\left({x}\mathrm{ln}\:{b}\right)\:={y} \\ $$$${x}=\frac{\mathrm{cosh}^{−\mathrm{1}} \:\frac{{y}}{\mathrm{2}}}{\mathrm{ln}\:{b}} \\ $$

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