"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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