Equation involving the logarithm Find the set of values of $k$ satisfying this equation for only one real root of $x.$
$$ \log(kx) = 2 \log(x+2)$$
I think that for the sake of satisfying the domain restriction:


*

*$ k \cdot x > 0 $

*$x+2 > 0$

*$ k \cdot x = (x+2)^2$

*$ b^2-2ac = 0$.


I am not sure what to do from here.

*

*$k⋅x>0$

*$x+2>0$ 

*$k⋅x=(x+2)^2$

*$b^2−2ac=0$



I am not sure what to do from here.

Well first thing is to correct 4.) $b^2-2ac = 0$ to $b^2 - 4ac=0$ (the $2$ was incorrect; it should be $4$) and then to write it in terms of $kx = (x+2)^2$
$kx = (x+2)^2$
$x^2 + (4-k)x + 4 = 0$ so $a = 1; b=4-k; c = 4$
so


*$(4-k)^2 - 16 = 0$.


Now just find the only values of $k$ where the following four are all true.


*

*$k⋅x>0$

*$x+2>0$ 

*$k⋅x=(x+2)^2$

*$(4-k)^2 - 16 = 0$


==== answer follows ======
So $(4-k)^2 = 16$
$4-k = \pm 4$
$k = 0, 8$.


*

*$kx > 0$


But if $k = 0$ then $k*x =0$.  So $k =0$ is impossible.  So $k = 8$.


*$kx = 8x = (x + 2)^2$


$8x = x^2 + 4x + 4$
$x^2 - 4x + 4 = 0$
$(x-2)^2 =0$ so 
$x = 2$


*$x + 2 = 2+ 2=4 > 0$


Yep.  That's fine.
So $k=8$ is the only value of $k$ which gives a single solution ($x =2$).
(If $k > 8$ then there will be two solutions and 4) fails (is greater than zero).  If $0< k < 8$ then 4) fails (is less than 0) and 3) has no solution.  If $k=0$ then 4) and 3) pass and there is a single solution to 3) but 1) and 2) fail so there is no overall solution.  If $k < 0$ then 4) fails (is greater than 0) and there are two solutions to 3) but 1) and 2) fail and there are no overall solution.)
