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Question Number 135172 by bemath last updated on 11/Mar/21

(x^2 −9)^(3x+5) = (x−3)^(x−1) .(x+3)^(x−1) Find solution

$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{3x}+\mathrm{5}} \:=\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} .\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{Find}\:\mathrm{solution} \\ $$

Answered by john_santu last updated on 11/Mar/21

⇒(x^2 −9)^(3x+5) = (x^2 −9)^(x−1) case(1) 3x+5 = x−1 ; x=−3 checking solution (0)^(−4) = ∞ so x=−3 no solution case(2) x^2 −9 = 1 ; x = ± (√(10)) solution case(3) x^2 −9 = 0 , x = 3 , since x=−3(rejected) case(4) x^2 −9 =−1 ; x = ±2(√2) cheking solution { (((−1)^(6(√2)+5) ≈ −0.0462)),(((−1)^((√2)−1) ≈ 0.266)) :} no solution { (((−1)^(−6(√2)+5) ≈ −0.0462)),(((−1)^(−(√2)−1) ≈ 0.266)) :}no solution ∴ the solution is { 3; ± (√(10)) }

$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{3}{x}+\mathrm{5}} \:=\:\left({x}^{\mathrm{2}} −\mathrm{9}\right)^{{x}−\mathrm{1}} \\ $$$${case}\left(\mathrm{1}\right)\:\mathrm{3}{x}+\mathrm{5}\:=\:{x}−\mathrm{1}\:;\:{x}=−\mathrm{3}\: \\ $$$${checking}\:{solution}\:\left(\mathrm{0}\right)^{−\mathrm{4}} \:=\:\infty\: \\ $$$${so}\:{x}=−\mathrm{3}\:{no}\:{solution} \\ $$$${case}\left(\mathrm{2}\right)\:{x}^{\mathrm{2}} −\mathrm{9}\:=\:\mathrm{1}\:;\:{x}\:=\:\pm\:\sqrt{\mathrm{10}}\:{solution} \\ $$$${case}\left(\mathrm{3}\right)\:{x}^{\mathrm{2}} −\mathrm{9}\:=\:\mathrm{0}\:,\:{x}\:=\:\mathrm{3}\:,\:{since}\:{x}=−\mathrm{3}\left({rejected}\right) \\ $$$${case}\left(\mathrm{4}\right)\:{x}^{\mathrm{2}} −\mathrm{9}\:=−\mathrm{1}\:;\:{x}\:=\:\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${cheking}\:{solution}\: \\ $$$$\begin{cases}{\left(−\mathrm{1}\right)^{\mathrm{6}\sqrt{\mathrm{2}}+\mathrm{5}} \:\approx\:−\mathrm{0}.\mathrm{0462}}\\{\left(−\mathrm{1}\right)^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\approx\:\mathrm{0}.\mathrm{266}}\end{cases}\:{no}\:{solution} \\ $$$$\begin{cases}{\left(−\mathrm{1}\right)^{−\mathrm{6}\sqrt{\mathrm{2}}+\mathrm{5}} \:\approx\:−\mathrm{0}.\mathrm{0462}}\\{\left(−\mathrm{1}\right)^{−\sqrt{\mathrm{2}}−\mathrm{1}} \:\approx\:\mathrm{0}.\mathrm{266}}\end{cases}{no}\:{solution} \\ $$$$\therefore\:{the}\:{solution}\:{is}\:\left\{\:\mathrm{3};\:\pm\:\sqrt{\mathrm{10}}\:\right\}\: \\ $$$$ \\ $$

Commented by metamorfose last updated on 11/Mar/21

why −3 isn′t a solution?...

$${why}\:−\mathrm{3}\:{isn}'{t}\:{a}\:{solution}?... \\ $$

Answered by metamorfose last updated on 11/Mar/21

(x^2 −9)^(3x+5) =(x^2 −9)^(x−1) ... if : (x^2 −9)=0 ... x=3 , x=−3 else : (3x+5)ln(x^2 −9)=(x−1)ln(x^2 −9) 2(x+3)ln(x^2 −9)=0 x=−3 , x^2 −9=1 x=−3 , x=±(√(10)) S_(IR) ={±3,±(√(10))}

$$\left({x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{3}{x}+\mathrm{5}} =\left({x}^{\mathrm{2}} −\mathrm{9}\right)^{{x}−\mathrm{1}} ... \\ $$$${if}\::\:\left({x}^{\mathrm{2}} −\mathrm{9}\right)=\mathrm{0}\:...\:{x}=\mathrm{3}\:,\:{x}=−\mathrm{3}\: \\ $$$${else}\:: \\ $$$$\left(\mathrm{3}{x}+\mathrm{5}\right){ln}\left({x}^{\mathrm{2}} −\mathrm{9}\right)=\left({x}−\mathrm{1}\right){ln}\left({x}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$$\mathrm{2}\left({x}+\mathrm{3}\right){ln}\left({x}^{\mathrm{2}} −\mathrm{9}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{3}\:,\:{x}^{\mathrm{2}} −\mathrm{9}=\mathrm{1} \\ $$$${x}=−\mathrm{3}\:,\:{x}=\pm\sqrt{\mathrm{10}} \\ $$$${S}_{{IR}} =\left\{\pm\mathrm{3},\pm\sqrt{\mathrm{10}}\right\} \\ $$

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