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Question Number 140730 by liberty last updated on 12/May/21
If equation 2log (x+3)=log ax  has only one solution. find the  value of a.
Ifequation2log(x+3)=logaxhasonlyonesolution.findthevalueofa.
Answered by mr W last updated on 12/May/21
a≠0  ax>0  x+3>0  (x+3)^2 =ax  x^2 +(6−a)x+9=0  Δ=(6−a)^2 −4×9=0  (6−a)^2 =36  6−a=±6  a=6±6=0, 12  valid is a=12
a0ax>0x+3>0(x+3)2=axx2+(6a)x+9=0Δ=(6a)24×9=0(6a)2=366a=±6a=6±6=0,12validisa=12
Commented by EDWIN88 last updated on 12/May/21
a = 12 or a< 0
a=12ora<0
Commented by EDWIN88 last updated on 12/May/21
put a = −1  2log (x+3) = log (−x)     { ((x+3>0⇒x>−3)),((−x>0⇒x<0)) :} ⇒−3<x<0  log (x+3)^2  = log (−x)  ⇒x^2 +6x+9+x = 0  ⇒x^2  +7x+9 =0   ⇒ x=((−7 +(√(13)))/2) ≈ −1.697  ⇒x = ((−7−(√(13)))/2) ≈−5.302 ← not solution
puta=12log(x+3)=log(x){x+3>0x>3x>0x<03<x<0log(x+3)2=log(x)x2+6x+9+x=0x2+7x+9=0x=7+1321.697x=71325.302notsolution
Commented by mr W last updated on 12/May/21
yes. you are right.
yes.youareright.

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