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


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

${(x^{2} - 9)}^{3 x + 5} = {(x - 3)}^{x - 1} . {(x + 3)}^{x - 1}$ $Find solution$
	Answered by john_santu last updated on 11/Mar/21
	$⇒(x^2 −9)^(3x+5) = (x^2 −9)^(x−1) case(1) 3x+5 = x−1 ; x=−3 checking solution (0)^(−4) = ∞ so x=−3 no solution case(2) x^2 −9 = 1 ; x = ± (√(10)) solution case(3) x^2 −9 = 0 , x = 3 , since x=−3(rejected) case(4) x^2 −9 =−1 ; x = ±2(√2) cheking solution { (((−1)^(6(√2)+5) ≈ −0.0462)),(((−1)^((√2)−1) ≈ 0.266)) :} no solution { (((−1)^(−6(√2)+5) ≈ −0.0462)),(((−1)^(−(√2)−1) ≈ 0.266)) :}no solution ∴ the solution is { 3; ± (√(10)) }$
	$\Rightarrow {(x^{2} - 9)}^{3 x + 5} = {(x^{2} - 9)}^{x - 1}$ $c a s e (1) 3 x + 5 = x - 1; x = - 3$ $c h e c k i n g s o l u t i o n {(0)}^{- 4} = \infty$ $s o x = - 3 n o s o l u t i o n$ $c a s e (2) x^{2} - 9 = 1; x = \pm \sqrt{10} s o l u t i o n$ $c a s e (3) x^{2} - 9 = 0, x = 3, s i n c e x = - 3 (r e j e c t e d)$ $c a s e (4) x^{2} - 9 = - 1; x = \pm 2 \sqrt{2}$ $c h e k i n g s o l u t i o n$ ${\begin{cases} {(- 1)}^{6 \sqrt{2} + 5} \approx - 0.0462 \\ {(- 1)}^{\sqrt{2} - 1} \approx 0.266 \end{cases} n o s o l u t i o n$ ${\begin{cases} {(- 1)}^{- 6 \sqrt{2} + 5} \approx - 0.0462 \\ {(- 1)}^{- \sqrt{2} - 1} \approx 0.266 \end{cases} n o s o l u t i o n$ $∴ t h e s o l u t i o n i s {3; \pm \sqrt{10}}$
	Commented by metamorfose last updated on 11/Mar/21

	$w h y - 3 {i s n}^{'} t a s o l u t i o n ? . . .$
	Answered by metamorfose last updated on 11/Mar/21
	$(x^2 −9)^(3x+5) =(x^2 −9)^(x−1) ... if : (x^2 −9)=0 ... x=3 , x=−3 else : (3x+5)ln(x^2 −9)=(x−1)ln(x^2 −9) 2(x+3)ln(x^2 −9)=0 x=−3 , x^2 −9=1 x=−3 , x=±(√(10)) S_(IR) ={±3,±(√(10))}$
	${(x^{2} - 9)}^{3 x + 5} = {(x^{2} - 9)}^{x - 1} . . .$ $i f : (x^{2} - 9) = 0 . . . x = 3, x = - 3$ $e l s e :$ $(3 x + 5) l n (x^{2} - 9) = (x - 1) l n (x^{2} - 9)$ $2 (x + 3) l n (x^{2} - 9) = 0$ $x = - 3, x^{2} - 9 = 1$ $x = - 3, x = \pm \sqrt{10}$ $S_{I R} = {\pm 3, \pm \sqrt{10}}$
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