Question

Let x,y be numbers in the interval (0,1) with the property that there exists a
positive number a different from 1 such that

Answer 1

\[\Huge {\log_xa+\log_ya=4\log_{xy}a}\]
\[\Huge{\text{prove that }x=y}\]

Answer 2

\[\large \log_xa+\log_ya=\log a\frac{ \log x+\log y }{ \log x \log y }=4\frac{ \log a }{\log xy }\]

Answer 3

\[ \large \frac{ \log xy }{ \log x \log y }=4\frac{ 1 }{ \log xy }\implies (\log xy)^2 =4\log x \log y\]

Answer 4

\[\large (\log xy)^2=\log^2x \log ^2y \implies \log xy=\log x \log y\]
consequently the only solution to this is x=y

Answer 5

not sure

Answer 6

\[x,y \in (0,1)\]

Answer 7

this part is not correct
\[ \large (\log xy)^2=\log^2x \log ^2y \implies \log xy=\log x \log y \]

Answer 8

should be like this
\[ (\log xy)^2 = (\log x)^2 + 2 \log x \log y + (\log y^2) = 4 \log x \log y \]
it's easy to deduce x=y from here. since log is bijective on \( (0, \infty) \)

Answer 9

*(log y)^2

Answer 10

experimentX is right. I think the easiest way to show it after the mistake would be \[(\log{xy})^2 = \log{xy}\cdot\log{xy} = 2\log{(x)}2\log{(y)} = \log{x^2}\log{y^2}\] Hence \[xy = x²; \; xy = y²;\; x,y \in (0,1) \Rightarrow x = y \]