"Find the real numbers $x$ and $y$, if $\lg^2\frac{x}{y}=3\cdot \lg\frac{x}{2024}\cdot \lg \frac{2024}{y}$."
[Wikipedia says that, according to the ISO 80000 specification(it costs a lot to read it, and time...), "$\lg$" should be the standard notation for the decimal logarithm $log_{10}.$ ]
*** The sign "***" means that the problem does not have a known author, as it appears in the MAGAZINE(aka REVISTE) Supliment cu Exerciții, September 2024; proposed for the 10th grade on page 9 with number S.L24.211.
ANSWER CiP
$$x=2024,\;\;y=2024$$
Solution CiP
Instead, we will solve the problem (...more than that, I don't venture to try...)
"Find the real numbers $x$ and $y$, if $\lg^2\frac{x}{y}=\lambda \cdot \lg\frac{x}{a} \cdot \lg \frac{a}{y}$"
$a$ and $\lambda$ being given positive real numbers, $\color {Red}{\lambda <4}$."
The ANSWER will be $\underline {x=a\;,\;y=a}$
The existence conditions of the problem, $\frac{x}{y}>0,\; \frac{x}{a}>0,\;\frac{a}{y}>0$ combined give $x>0$ and $y>0$.
I will use the inequality
$$4\cdot u \cdot v \leqslant (u+v)^2 \tag{1}$$
with the sign $"="$ in (1) if and only if $u=v$. Let $A:=\lg\frac{x}{y}$;
$$A^2\underset{eq}{=}\lambda \cdot \lg\frac{x}{a}\cdot \lg\frac{a}{y}=\frac{\lambda}{4}\cdot 4\lg\frac{x}{a}\lg\frac{a}{y}\;\;\;\;\overset{(1)}{\underset{u=\lg\frac{x}{a}\;v=\lg\frac{a}{y}}{\leqslant}}\; \;\;\;\frac{\lambda}{4} \cdot \left ( \lg\frac{x}{a}+\lg \frac{a}{y} \right)^2 =$$
$$=\frac{\lambda}{4}\cdot \left [\lg \left (\frac{x}{a}\cdot \frac{a}{y}\right )\right ]^2=\frac{\lambda}{4}\cdot \lg^2 \frac{x}{y}=\frac{\lambda}{4}\cdot A^2,$$
hence $A^2 \leqslant \frac{\lambda}{4} \cdot A^2\;\Leftrightarrow\;A^2\cdot (1-\frac{\lambda}{4})\leqslant 0$ but which in the given condition $\lambda <4$ implies $A=0.$
I got $\lg\frac{x}{y}=0$ so $x=y$ and, from the equation, that one of the conditions $\lg \frac{x}{a}=0$ or $\lg \frac{a}{y}=0$ occurs, so $x=a$ or $y=a$, (!)actually both.
$\blacksquare$
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