Equation involving the logarithm

Question

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:

1. $ k \cdot x > 0 $
2. $x+2 > 0$
3. $ k \cdot x = (x+2)^2$
4. $ b^2-2ac = 0$.

I am not sure what to do from here.

Answer 1

Find the set of values of $k$ satisfying
\begin{align} \log(kx) &= 2 \log(x+2) \tag{1}\label{1} \end{align} for only one real root of $x$.

Equation \begin{align} k \cdot x &= (x+2)^2 \tag{2}\label{2} \end{align} follows from \eqref{1} and can be transformed as

\begin{align} (x+2)^2 -kx&=0 \\ (x+2)^2 -k(x+2)+2k&=0 \end{align}

with the roots \begin{align} x+2&=\tfrac12\,k\pm\tfrac12\sqrt{k^2-8k} \tag{3}\label{3} . \end{align}

The domain restrictions

\begin{align} kx&>0 \tag{4}\label{4} ,\\ x+2&>0 \tag{5}\label{5} \end{align}

suggest that we need to consider two cases:

\begin{align} \text{Case 1. }\quad k&>0,\quad x>0 \tag{6}\label{6} ,\\ \text{Case 2. }\quad k&<0,\quad x\in(-2,0) \tag{7}\label{7} . \end{align}

In the first case there is only one value ($k=8$) that is useful for \eqref{1}.

In the second case, when $k<0,\quad x\in(-2,0)$, since the left-hand side of \eqref{3} is positive, we need to check, for which values of negative $k$ \begin{align} x+2&=\tfrac12\,k+\tfrac12\sqrt{k^2-8k} \tag{8}\label{8} \end{align} holds.

From equation \eqref{8} and condition \eqref{7},

\begin{align} 0&<\tfrac12\,k+\tfrac12\sqrt{k^2-8k} <2 \tag{9}\label{9} \end{align}

Let $\kappa=-k$, $\kappa>0$. Then condition
\begin{align} 0&<-\tfrac12\,\kappa+\tfrac12\sqrt{\kappa^2+8\kappa} \end{align} always holds for any $\kappa>0$, that is, for any $k<0$.

Next, consider

\begin{align} -\tfrac12\,\kappa+\tfrac12\,\kappa\,\sqrt{1+\frac8\kappa} <2 ,\\ -\kappa+\kappa\,\sqrt{1+\frac8\kappa} <4 ,\\ \kappa\,\sqrt{1+\frac8\kappa} <4+\kappa ,\\ \kappa^2+8\kappa <16+8\kappa+\kappa^2 , \end{align} which also holds for all $\kappa>0$, that is, for all $k<0$ .

Summarizing, the answer is: \begin{align} k\in(-\infty,0)\cup\{8\} . \end{align}

Illustration for the Case 1:

Illustration for the Case 2:

Answer 2

1. $k⋅x>0$
2. $x+2>0$
3. $k⋅x=(x+2)^2$
4. $b^2−2ac=0$

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

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

1. $k⋅x>0$
2. $x+2>0$
3. $k⋅x=(x+2)^2$
4. $(4-k)^2 - 16 = 0$

==== answer follows ======

So $(4-k)^2 = 16$

$4-k = \pm 4$

$k = 0, 8$.

1. $kx > 0$

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

3. $kx = 8x = (x + 2)^2$

$8x = x^2 + 4x + 4$

$x^2 - 4x + 4 = 0$

$(x-2)^2 =0$ so

$x = 2$

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.)