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Question Number 158255 by daus last updated on 01/Nov/21

Commented by daus last updated on 01/Nov/21

solve the x

solvethex

Commented by cortano last updated on 01/Nov/21

((2/3))^(2/x) −((2/3))^(1/x) −1=0 ((2/3))^(1/x) = q ⇒q^2 −q−1=0 ⇒q=((1+(√5))/2) , ((2/3))^(1/x) =((1+(√5))/2) ⇒(1/x)= log _(2/3) (((1+(√5))/2)) ⇒x= log _((((1+(√5))/2))) ((2/3))

(23)2x−(23)1x−1=0(23)1x=q⇒q2−q−1=0⇒q=1+52,(23)1x=1+52⇒1x=log23(1+52)⇒x=log(1+52)(23)

Answered by MathsFan last updated on 01/Nov/21

(4^(1/x) /4^(1/x) )−(6^(1/x) /4^(1/x) )=(9^(1/x) /4^(1/x) ) 1−((3^(1/x) •2^(1/x) )/(2^(1/x) •2^(1/x) ))=(3^((1/x)•2) /2^((1/x)•2) ) 1−((3^(1/x) /2^(1/x) ))=((3^(1/x) /2^(1/x) ))^2 ((3^(1/x) /2^(1/x) ))^2 +((3^(1/x) /2^(1/x) ))=1 ((3^(1/x) /2^(1/x) ))^2 +((3^(1/x) /2^(1/x) ))+(1/4)=1+(1/4) [((3^(1/x) /2^(1/x) ))+(1/2)]^2 =(5/4) ((3^(1/x) /2^(1/x) ))=((±(√5))/2)−(1/2) ln((3/2))^(1/x) =ln(((√5)−1)/2) (1/x)=((ln(((√5)−1)/2))/(ln((3/2)))) x_1 =((ln((3/2)))/(ln(((√5)−1)/2))) x_2 =((ln((3/2)))/(ln((−(√5)−1)/2)))

41x41x−61x41x=91x41x1−31x∙21x21x∙21x=31x∙221x∙21−(31x21x)=(31x21x)2(31x21x)2+(31x21x)=1(31x21x)2+(31x21x)+14=1+14[(31x21x)+12]2=54(31x21x)=±52−12ln(32)1x=ln5−121x=ln5−12ln(32)x1=ln(32)ln5−12x2=ln(32)ln−5−12

Answered by yeti123 last updated on 01/Nov/21

4^(1/x) − 6^(1/x) = 9^(1/x) ((4/6))^(1/x) − 1 = ((9/6))^(1/x) ((2/3))^(1/x) − 1 − ((3/2))^(1/x) = 0 y = (2/3)^(1/x) ⇒ y − 1 − (1/y) = 0 y^2 − y − 1 = 0 ⋮

41/x−61/x=91/x(46)1/x−1=(96)1/x(23)1/x−1−(32)1/x=0y=(2/3)1/x⇒y−1−1y=0y2−y−1=0⋮

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