The Power of Logs!

The first thing to be noted is that if $a,b,c,d$ are in GP, then $\log { a, } \log { b, } \log { c, } \log { d } \quad$ are in arithmetic progression. (you can check it yourself). Therefore if $x,{ x }^{ \log _{ 10 }{ x } },{ y }^{ \log _{ 10 }{ y } },{ (xy) }^{ \log _{ 10 }{ (xy) }}$ are in GP, then , $\log _{ 10 }{ (x) } ,\log _{ 10 }{ ({ x }^{ \log _{ 10 }{ x } }) } ,\log _{ 10 }{ ({ y }^{ \log _{ 10 }{ y } }) } ,\log _{ 10 }{ ({ (xy) }^{ \log _{ 10 }{ (xy) } }) }$

or simply, $\\ \log { x } ,{ (\log { x } })^{ 2 }{ ,(\log { y } })^{ 2 },{ (\log { (xy) } ) }^{ 2 }\\$ are in AP. Therefore, $({ \log { (xy)) } }^{ 2 }\quad -\quad ({ \log { y } })^{ 2 }\quad =\quad { (\log { y) } }^{ 2 }\quad -\quad ({ \log { x } })^{ 2 }\quad (=\quad common\quad difference\quad in\quad the\quad AP)\\ \Longrightarrow \quad (\log { xy\quad + } \log { y\quad )( } \log { xy\quad - } \log { y } )\quad =\quad { (\log { y) } }^{ 2 }\quad -\quad ({ \log { x } })^{ 2 }\\ \Longrightarrow \quad (\log { x } +\quad 2\log { y } )(\log { x } )\quad =\quad { (\log { y) } }^{ 2 }\quad -\quad ({ \log { x } })^{ 2 }\\ \Longrightarrow \quad ({ \log { y } ) }^{ 2 }\quad -2\log { x } \log { y } -2({ \log { x) } }^{ 2 }\quad =\quad 0\quad .......(1)$

Similarly ,

$\\ \\ ({ \log { y } ) }^{ 2 }-({ \log { x) } }^{ 2 }\quad =\quad { (\log { x } ) }^{ 2 }\quad -\quad \log { x } \\ \Longrightarrow \quad ({ \log { y } ) }^{ 2 }-2({ \log { x) } }^{ 2 }+\log { x } \quad =\quad 0 ......(2)$

$\\ (2)\quad -\quad (1)\\ \\ \Longrightarrow 2\log { y } \log { x } +\log { x } \quad =\quad 0 .....(3)\\ \Longrightarrow \quad \log { x=0\quad or } \log { y\quad =\quad -\frac { 1 }{ 2 } } \\ \Longrightarrow x=1\quad or\quad y=\quad \frac { 1 }{ \sqrt { 10 } } \\$

Case 1:- When $x=1$ ,equation $(3) or (1) or (2)$ gives us $\log { y\quad =\quad 0\quad or\quad y=\quad 1 }$ hence one solution is $\\ { \boxed { (1,1) } }$

Case 2 :- When $y=\frac { 1 }{ \sqrt { 10 } }\quad or\quad \log { y } \quad =\quad -\frac { 1 }{ 2 }$ ,equation $(2)$ gives us,

$\\ 2{ (\log { x } ) }^{ 2 }-\quad \log { x } -\frac { 1 }{ 4 } \quad =0\quad \\$

which is a quadratic equation in log x.

On solving by quadratic formula,

$\\ \log { x } =\frac { 1\pm \sqrt { 3 } }{ 4 } \\$ or $x={ 10 }^{ \frac { 1+\sqrt { 3 } }{ 4 } },{ 10 }^{ \frac { 1-\sqrt { 3 } }{ 4 } }\\ \\$

which gives us two more solutions $\boxed{({ 10 }^{ \frac { 1+\sqrt { 3 } }{ 4 } },\frac { 1 }{ \sqrt { 10 } } ),({ 10 }^{ \frac { 1-\sqrt { 3 } }{ 4 } },\frac { 1 }{ \sqrt { 10 } } )}$

Hence a total of $\boxed { 3 }$ ordered pairs.

$Note :-$ All logarithms used above are in base 10.

Since $x, x^{\log{ x}}, y^{\log{ y}},(xy)^{\log(xy)}$ are in a geometric progression, we can take the log of every term to get an arithmetic progression. Thus, $\log { x},(\log x)^2,(\log y)^2$ and $(\log(xy))^2$ are in an arithmetic progression.

From the first two terms, we can see that the common difference is $(\log x)^2-\log{ x}$

Then, from the last two terms, we have:

$(\log (xy))^2-(\log y)^2 = (\log x)^2 - \log{ x}$

$(\log{ x} + \log{ y})^2-(\log y)^2 = (\log x)^2 - \log{ x}$

$(\log x)^2+2\log{ x}\log{ y}+(\log y)^2-(\log y)^2 = (\log x)^2 - \log{ x}$

$2\log{ x}\log{ y}=-\log{ x}$

$(2\log{ y}+1)\log{ x}=0$

So either $\log{ y}=-\frac{1}{2}$ or $\log{ x}=0$ .

If $\log{ x}=0$ , we have the first two terms of the arithmetic progression are $0$ , so the third term must also be $0$ , meaning $\log{ y}=0$ , giving an ordered pair $(1,1)$

If $\log{ y}=-\frac{1}{2}$ , then we have that $\log { x},(\log x)^2$ ,and $(\log y)^2$ form an arithmetic progression.

$(\log{ y})^2-(\log{ x})^2=(\log{ x})^2-\log{ x}$

$\frac{1}{4}=2(\log{ x})^2-\log{ x}$

$8(\log{ x})^2-\log{ x}-1=0$

Since the discriminant is positive, there are $2$ real solutions for $\log{ x}$ , making $2$ more pairs $(x,y)$

Thus, there are a total of $\boxed{3}$ ordered pairs $(x,y)$ that satisfy the equation.