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Question Number 118651 by Sherjon last updated on 18/Oct/20

Commented by Sherjon last updated on 18/Oct/20

x_1 +x_2 +...=? Why

x1+x2+...=?Why

Commented by bramlexs22 last updated on 19/Oct/20

log _2 (9^(x−1) +7)=log _2 (4.3^(x−1) +4) ⇒ (3^(x−1) )^2 −4(3^(x−1) )+3 = 0 have the roots are → { (3^(x_1 −1) ),(3^(x_2 −1) ) :} By the Vieta′s ⇒ 3^(x_1 −1) × 3^(x_2 −1) = 3 ⇒3^(x_1 +x_2 −2) = 3^1 ⇒x_1 +x_2 = 3

log2(9x−1+7)=log2(4.3x−1+4)⇒(3x−1)2−4(3x−1)+3=0havetherootsare→{3x1−13x2−1BytheVieta′s⇒3x1−1×3x2−1=3⇒3x1+x2−2=31⇒x1+x2=3

Answered by Dwaipayan Shikari last updated on 18/Oct/20

log_2 (9^(x−1) +7)−log_2 (3^(x−1) +1)=2 ((9^(x−1) +7)/(3^(x−1) +1))=4 3^(2(x−1)) +7=4.3^(x−1) +4 a^2 −4a+3=0 (3^(x−1) =a) a=1 or a=3 3^(x−1) =1⇒x−1=0⇒x=1 Or 3^(x−1) =3^1 ⇒x=2 So x_1 +x_2 =3

log2(9x−1+7)−log2(3x−1+1)=29x−1+73x−1+1=432(x−1)+7=4.3x−1+4a2−4a+3=0(3x−1=a)a=1ora=33x−1=1⇒x−1=0⇒x=1Or3x−1=31⇒x=2Sox1+x2=3

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