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Question Number 151205 by EDWIN88 last updated on 19/Aug/21

   determinant ((((2+(√3))^x +1 =(2(√(2+(√3))))^x )),((x =? )))

$$\underbrace{ }\:\begin{array}{|c|c|}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} +\mathrm{1}\:=\left(\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} }\\{{x}\:=?\:}\\\hline\end{array} \\ $$

Answered by bramlexs22 last updated on 19/Aug/21

 (1/((2−(√3))^x )) + 1 = (2^x /((2−(√3))^(x/2) ))    (1/((2−(√3))^(x/2) )) + (2−(√3))^(x/2)  = 2^x   (1/((2−(√3))^x )) +(2−(√3))^x  = 4^x −2   (2+(√3))^x  + (2−(√3))^x  = 4^x −2  x = 2 , check ⇒LHS≡(2+(√3))^2 +(2−(√3))^2  = 7+7=14  RHS ≡ 4^2 −2=14

$$\:\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} }\:+\:\mathrm{1}\:=\:\frac{\mathrm{2}^{\mathrm{x}} }{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{x}}{\mathrm{2}}} }\: \\ $$$$\:\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{x}}{\mathrm{2}}} }\:+\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{x}}{\mathrm{2}}} \:=\:\mathrm{2}^{\mathrm{x}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} }\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:=\:\mathrm{4}^{\mathrm{x}} −\mathrm{2} \\ $$$$\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:+\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:=\:\mathrm{4}^{\mathrm{x}} −\mathrm{2} \\ $$$$\mathrm{x}\:=\:\mathrm{2}\:,\:\mathrm{check}\:\Rightarrow\mathrm{LHS}\equiv\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:=\:\mathrm{7}+\mathrm{7}=\mathrm{14} \\ $$$$\mathrm{RHS}\:\equiv\:\mathrm{4}^{\mathrm{2}} −\mathrm{2}=\mathrm{14}\: \\ $$

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