Menu Close

If-equation-2log-x-3-log-ax-has-only-one-solution-find-the-value-of-a-




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.
$$\mathrm{If}\:\mathrm{equation}\:\mathrm{2log}\:\left(\mathrm{x}+\mathrm{3}\right)=\mathrm{log}\:\mathrm{ax} \\ $$$$\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}.\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{a}. \\ $$
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
$${a}\neq\mathrm{0} \\ $$$${ax}>\mathrm{0} \\ $$$${x}+\mathrm{3}>\mathrm{0} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} ={ax} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{6}−{a}\right){x}+\mathrm{9}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{6}−{a}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{9}=\mathrm{0} \\ $$$$\left(\mathrm{6}−{a}\right)^{\mathrm{2}} =\mathrm{36} \\ $$$$\mathrm{6}−{a}=\pm\mathrm{6} \\ $$$${a}=\mathrm{6}\pm\mathrm{6}=\mathrm{0},\:\mathrm{12} \\ $$$${valid}\:{is}\:{a}=\mathrm{12} \\ $$
Commented by EDWIN88 last updated on 12/May/21
a = 12 or a< 0
$$\mathrm{a}\:=\:\mathrm{12}\:\mathrm{or}\:\mathrm{a}<\:\mathrm{0} \\ $$
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
$$\mathrm{put}\:\mathrm{a}\:=\:−\mathrm{1} \\ $$$$\mathrm{2log}\:\left(\mathrm{x}+\mathrm{3}\right)\:=\:\mathrm{log}\:\left(−\mathrm{x}\right)\: \\ $$$$\:\begin{cases}{\mathrm{x}+\mathrm{3}>\mathrm{0}\Rightarrow\mathrm{x}>−\mathrm{3}}\\{−\mathrm{x}>\mathrm{0}\Rightarrow\mathrm{x}<\mathrm{0}}\end{cases}\:\Rightarrow−\mathrm{3}<\mathrm{x}<\mathrm{0} \\ $$$$\mathrm{log}\:\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{log}\:\left(−\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{9}+\mathrm{x}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \:+\mathrm{7x}+\mathrm{9}\:=\mathrm{0}\: \\ $$$$\Rightarrow\:\mathrm{x}=\frac{−\mathrm{7}\:+\sqrt{\mathrm{13}}}{\mathrm{2}}\:\approx\:−\mathrm{1}.\mathrm{697} \\ $$$$\Rightarrow\mathrm{x}\:=\:\frac{−\mathrm{7}−\sqrt{\mathrm{13}}}{\mathrm{2}}\:\approx−\mathrm{5}.\mathrm{302}\:\leftarrow\:\mathrm{not}\:\mathrm{solution} \\ $$
Commented by mr W last updated on 12/May/21
yes. you are right.
$${yes}.\:{you}\:{are}\:{right}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *