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Question Number 140730 by liberty last updated on 12/May/21

$$\mathrm{If}\mathrm{equation}2\log (\mathrm{x}+3)=\log \mathrm{ax}$$$$\mathrm{has}\mathrm{only}\mathrm{one}\mathrm{solution}.\mathrm{find}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\mathrm{a}.$$

Answered by mr W last updated on 12/May/21

$$a\ne 0$$$$ax>0$$$$x+3>0$$$${(x+3)}^2=ax$$$$x^2+(6-a)x+9=0$$$$\mathrm{\Delta }={(6-a)}^2-4\times 9=0$$$${(6-a)}^2=36$$$$6-a=\pm 6$$$$a=6\pm 6=0,12$$$$validisa=12$$

Commented by EDWIN88 last updated on 12/May/21

$$\mathrm{a}=12\mathrm{or}\mathrm{a}<0$$

Commented by EDWIN88 last updated on 12/May/21

$$\mathrm{put}\mathrm{a}=-1$$$$2\log (\mathrm{x}+3)=\log (-\mathrm{x})$$$$\{\begin{array}{l}\mathrm{x}+3>0\Rightarrow \mathrm{x}>-3\\ -\mathrm{x}>0\Rightarrow \mathrm{x}<0\end{array}\Rightarrow -3<\mathrm{x}<0$$$$\log {(\mathrm{x}+3)}^2=\log (-\mathrm{x})$$$$\Rightarrow {\mathrm{x}}^2+6\mathrm{x}+9+\mathrm{x}=0$$$$\Rightarrow {\mathrm{x}}^2+7\mathrm{x}+9=0$$$$\Rightarrow \mathrm{x}=\frac{-7+\sqrt{13}}{2}\approx -1.697$$$$\Rightarrow \mathrm{x}=\frac{-7-\sqrt{13}}{2}\approx -5.302\leftarrow \mathrm{not}\mathrm{solution}$$

Commented by mr W last updated on 12/May/21

$$yes.youareright.$$

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