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Question Number 9543 by Joel575 last updated on 14/Dec/16

(2−(√3))^x + (7−4(√3))(2+(√3))^x = 4(2−(√3)), x ≠ 0 What is the value of x ?

$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}} \:+\:\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} \:=\:\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right),\:{x}\:\neq\:\mathrm{0} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:?\: \\ $$

Answered by mrW last updated on 14/Dec/16

u=(2−(√3))^x (2+(√3))^x =(1/((2−(√3))^x ))=(1/u) ⇒u+((7−4(√3))/u)=4(2−(√3)) u^2 −4(2−(√3))u+(7−4(√3))=0 u=((4(2−(√3))±(√(16(2−(√3))^2 −4(7−4(√3)))))/2) u=((4(2−(√3))±2(2(√3)−3))/2) u_1 =4−2(√3)+2(√3)−3=1 u_2 =4−2(√3)−2(√3)+3=7−4(√3)=(2−(√3))^2 (2−(√3))^x =u_1 =1 ⇒x_1 =0 (2−(√3))^x =u_2 =(2−(√3))^2 ⇒x_2 =2

$$\mathrm{u}=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} =\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} }=\frac{\mathrm{1}}{\mathrm{u}} \\ $$$$\Rightarrow\mathrm{u}+\frac{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{u}}=\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\mathrm{u}+\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\mathrm{u}=\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\pm\sqrt{\mathrm{16}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)}}{\mathrm{2}} \\ $$$$\mathrm{u}=\frac{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\pm\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right)}{\mathrm{2}} \\ $$$$\mathrm{u}_{\mathrm{1}} =\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}=\mathrm{1} \\ $$$$\mathrm{u}_{\mathrm{2}} =\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}=\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} =\mathrm{u}_{\mathrm{1}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{1}} =\mathrm{0} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} =\mathrm{u}_{\mathrm{2}} =\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{2}} =\mathrm{2} \\ $$

Commented by Joel575 last updated on 14/Dec/16

Can u explain to me why (2+(√3))^x =(1/((2−(√3))^x )) ?

$$\mathrm{Can}\:\mathrm{u}\:\mathrm{explain}\:\mathrm{to}\:\mathrm{me}\:\mathrm{why} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} =\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} }\:\:? \\ $$

Commented by tawakalitu last updated on 14/Dec/16

Rationalize: (2 + (√3)), you will get. (((2 + (√3)))/1) × (((2 − (√3)))/((2 − (√3)))) = ((4 − (√3) + (√3) − 3)/((2 − (√3)))) = ((4 − 0 − 3)/((2 − (√3)))) = (1/((2 − (√3)))) Now, (2 + (√3))^x = ((1/(2 − (√3))))^x Now, (2 + (√3))^x = (1^x /((2 − (√3))^x )) Now, (2 + (√3))^x = (1/((2 − (√3))^x )) That is how.

$$\mathrm{Rationalize}:\:\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right),\:\mathrm{you}\:\mathrm{will}\:\mathrm{get}. \\ $$$$\frac{\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)}{\mathrm{1}}\:×\:\frac{\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)}{\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)} \\ $$$$=\:\frac{\mathrm{4}\:−\:\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{3}}\:−\:\mathrm{3}}{\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)} \\ $$$$=\:\frac{\mathrm{4}\:−\:\mathrm{0}\:−\:\mathrm{3}}{\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{Now},\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:=\:\left(\frac{\mathrm{1}}{\mathrm{2}\:−\:\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} \\ $$$$\mathrm{Now},\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:=\:\frac{\mathrm{1}^{\mathrm{x}} }{\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} } \\ $$$$\mathrm{Now},\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:=\:\frac{\mathrm{1}}{\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} } \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{how}. \\ $$

Commented by mrW last updated on 14/Dec/16

(2+(√3))(2−(√3))=2^2 −((√3))^2 =1 2+(√3)=(1/(2−(√3))) (2+(√3))^x =((1/(2−(√3))))^x =(1/((2−(√3))^x ))

$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=\mathrm{2}^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}+\sqrt{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} =\left(\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} =\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} } \\ $$

Commented by Joel575 last updated on 15/Dec/16

ok. thank you very much

$$\mathrm{ok}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

Answered by mrW last updated on 15/Dec/16

it can be simplified: (7−4(√3))=(2−(√3))^2 (2+(√3))^x =(2−(√3))^(−x) (2−(√3))^x +(2−(√3))^(2−x) =4(2−(√3)) (2−(√3))^(x−1) +(2−(√3))^(1−x) =4 u=(2−(√3))^(x−1) ⇒u+(1/u)=4 u^2 −4u+1=0 u^2 −4u+4−((√3))^2 =0 (u−2+(√3))(u−2−(√3))=0 u_1 =2−(√3) u_2 =2+(√3)=(2−(√3))^(−1) (2−(√3))^(x−1) =u_1 =2−(√3) x−1=1 ⇒x=2 (2−(√3))^(x−1) =u_2 =(2−(√3))^(−1) x−1=−1 ⇒x=0

$$\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{simplified}: \\ $$$$\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} =\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{−\mathrm{x}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}} +\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}−\mathrm{x}} =\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{1}} +\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{1}−\mathrm{x}} =\mathrm{4} \\ $$$$\mathrm{u}=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{u}+\frac{\mathrm{1}}{\mathrm{u}}=\mathrm{4} \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{4u}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{4u}+\mathrm{4}−\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{u}−\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{u}−\mathrm{2}−\sqrt{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\mathrm{u}_{\mathrm{1}} =\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\mathrm{u}_{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}}=\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{−\mathrm{1}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{1}} =\mathrm{u}_{\mathrm{1}} =\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\mathrm{x}−\mathrm{1}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{1}} =\mathrm{u}_{\mathrm{2}} =\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{−\mathrm{1}} \\ $$$$\mathrm{x}−\mathrm{1}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{0} \\ $$

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