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Question Number 64916 by Masumsiddiqui399@gmail.com last updated on 23/Jul/19

Commented by Tawa1 last updated on 23/Jul/19

(2 + (√3))^x  + ((1/(2 + (√3))))^x  =  4               [since    2 − (√3)  =  (1/(2 + (√3)))  on rationalise]  (2 + (√3))^x  + (1/((2 + (√3))^x )) =  4  Let     (2 + (√3))^x   =  y     y + (1/y) =  4     y^2  + 1 =  4y     y^2  − 4y + 1   =  0  a = 1, b = − 4, c = 1     ⇒  y  =   ((− (− 4) ± (√((− 4)^2  − 4(1)(1))))/(2(1)))     ⇒  y  =   ((4 ± (√(16− 4)))/2)     ⇒  y  =   ((4 ± (√(12)))/2)     ⇒  y  =   ((4 ± (√(4 × 3)))/2)     ⇒  y  =   ((4 ± 2(√3))/2)     ⇒  y  =   2 ± (√3)     ⇒  y  =   2 + (√3)    or   y  =  2 − (√3)  Recall the  Let     (2 + (√3))^x   =  y  ⇒       (2 + (√3))^x   =  (2 + (√3))^1   Base are the same:     ⇒       x = 1  Or       (2 + (√3))^x   =  (2 − (√3))^1        (2 + (√3))^x   =  (1/(2 + (√3)))       (2 + (√3))^x   =  (2 + (√3))^(−1)   Base are the same:     ⇒       x = − 1  Finally                     x  =  1     or    x = − 1

(2+3)x+(12+3)x=4[since23=12+3onrationalise](2+3)x+1(2+3)x=4Let(2+3)x=yy+1y=4y2+1=4yy24y+1=0a=1,b=4,c=1y=(4)±(4)24(1)(1)2(1)y=4±1642y=4±122y=4±4×32y=4±232y=2±3y=2+3ory=23RecalltheLet(2+3)x=y(2+3)x=(2+3)1Basearethesame:x=1Or(2+3)x=(23)1(2+3)x=12+3(2+3)x=(2+3)1Basearethesame:x=1Finallyx=1orx=1

Commented by Tawa1 last updated on 23/Jul/19

Hahahaha,  i solve like sir  MrW,  sir  Mjs ,   sir  Tanmay,  sir Ajfour,  Am learning everyday here.  God bless everybody here.

Hahahaha,isolvelikesirMrW,sirMjs,sirTanmay,sirAjfour,Amlearningeverydayhere.Godblesseverybodyhere.

Commented by MJS last updated on 23/Jul/19

good!

good!

Commented by mathmax by abdo last updated on 23/Jul/19

let a =2+(√3)   we have 2−(√3)=(1/(2+(√3)))  so (e)⇒a^x  +(1/a^x ) =4 ⇒  a^(2x)  +1 =4a^x  ⇒a^(2x) −4a^x  +1 =0 ⇒t^2 −4t +1 =0  with t=a^x   Δ^′  =4−1 =3 ⇒t_1 =2+(√3)  and t_2 =2−(√3)  a^x =2+(√3) ⇒(2+(√3))^x  =2+(√3) ⇒(2+(√3))^(x−1)  =1 ⇒x−1=0 ⇒x=1  a^x =2−(√3)⇒a^x  =(2+(√3))^(−1)  ⇒(2+(√3))^(x+1) =1 ⇒x+1 =0⇒x=−1

leta=2+3wehave23=12+3so(e)ax+1ax=4a2x+1=4axa2x4ax+1=0t24t+1=0witht=axΔ=41=3t1=2+3andt2=23ax=2+3(2+3)x=2+3(2+3)x1=1x1=0x=1ax=23ax=(2+3)1(2+3)x+1=1x+1=0x=1

Commented by Masumsiddiqui399@gmail.com last updated on 23/Jul/19

thank u sir

thankusir

Commented by Masumsiddiqui399@gmail.com last updated on 23/Jul/19

thank u sir

thankusir

Answered by MJS last updated on 23/Jul/19

(2+(√3))^x +(2−(√3))^x =4  obviously x_2 =1  2+(√3)+2−(√3)=4  in this case  (2+(√3))^(−1) =2−(√3)  ⇒  (2+(√3))^x +(2−(√3))^x =(2+(√3))^(−x) +(2−(√3))^(−x)   ⇒ x_2 =−1

(2+3)x+(23)x=4obviouslyx2=12+3+23=4inthiscase(2+3)1=23(2+3)x+(23)x=(2+3)x+(23)xx2=1

Commented by MJS last updated on 23/Jul/19

I think generally we can only try  (a+(√b))^x +(a−(√b))^x =y    if y=2a ⇒ x_1 =1    if (a+(√b))^(−1) =a−(√b) ⇔ b=a^2 −1  (a+(√(a^2 −1)))^x +(1/((a+(√(a^2 −1)))^x ))=y  2cosh (xln (a+(√(a^2 −1))))=y  ⇒ x=±((arcosh (y/2))/(ln (a+(√(a^2 −1)))))

Ithinkgenerallywecanonlytry(a+b)x+(ab)x=yify=2ax1=1if(a+b)1=abb=a21(a+a21)x+1(a+a21)x=y2cosh(xln(a+a21))=yx=±arcoshy2ln(a+a21)

Commented by Tawa1 last updated on 23/Jul/19

Wow, God bless you sir.  I will note this too.

Wow,Godblessyousir.Iwillnotethistoo.

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