Task B-3.4. Determine all values of the positive real number $k$ for which the equation

$$
\frac{\log (k x)}{\log (x+1)}=2
$$

has exactly one solution.
## Solution.

First, let's write down the conditions for which the given equation makes sense.

1. $x+1>0, x>-1$;
2. $k x>0$, and since $k>0$ it follows that $x>0$;
3. $\log (x+1) \neq 0, x \neq 0$.

From conditions (1), (2), and (3), we get $x>0$.

(1 point)

The given equation can be written in the form

$$
\begin{aligned}
\log (k x) & =2 \log (x+1) \\
& =\log (x+1)^{2}
\end{aligned}
$$

Then we have

$$
\begin{aligned}
& k x=(x+1)^{2} \\
& x^{2}-x(k-2)+1=0
\end{aligned}
$$

This equation has exactly one solution if its discriminant is zero, i.e.,

$$
(k-2)^{2}-4=0
$$

Since $k \neq 0$, the only solution is $k=4$.