Some Logs These Are!

Let $z:=\log_y x$ , then $\log_x y=\frac1z$ and so $z+\frac1z=-2$ , giving us $(z+1)^2=0$ , hence $z=-1$ . Therefore $\log_y x=-1$ , meaning that $y^{-1}=x$ . So we get $xy=y^{-1}y=\boxed{1}$ .

Using the Properties of Logarithms , $\log_{y}{x}+\log_{x}{y}=-2 \color{#D61F06} \Rightarrow \color{#333333} \frac{\log_{}{x}}{\log_{}{y}}+\frac{\log_{}{y}}{\log_{}{x}}=-2$ $\color{#D61F06} \Rightarrow \color{#333333} \frac{(\log_{}{x})^{2}+(\log_{}{y})^{2}}{\log_{}{x}\log_{}{y}}=-2\color{#D61F06} \Rightarrow \color{#333333} (\log_{}{x})^{2}+(\log_{}{y})^{2}=-2\log_{}{x}\log_{}{y}$ $\color{#D61F06} \Rightarrow \color{#333333} (\log_{}{x})^{2}+2\log_{}{x}\log_{}{y}+(\log_{}{y})^{2}=0\color{#D61F06} \Rightarrow \color{#333333} (\log_{}{x}+\log_{}{y})^{2}=0$ $\color{#D61F06} \Rightarrow \color{#333333} \log_{}{x}+\log_{}{y}=0\color{#D61F06} \Rightarrow \color{#333333} \log_{}{x}=-\log_{}{y} \color{#D61F06} \Rightarrow \color{#333333} \log_{}{x}=\log_{}{\frac{1}{y}}$ $a>0, \space a \neq 1:\space\space\color{#D61F06} a^{x}=a^{y} \color{#333333} \Leftrightarrow \color{#D61F06} x=y$ $\color{#D61F06} \Rightarrow \color{#333333} 10^{\log_{}{x}}=10^{\log_{}{\frac{1}{y}}} \color{#D61F06} \Rightarrow \color{#333333} x=\frac{1}{y}$ $\color{#D61F06} \Rightarrow \color{#333333} xy=\color{#3D99F6}{1} \color{#333333} \space\space\space\Box$

I like to take it step by step and apply the log rules.

$log_y x + log_x y = -2 \\ \Rightarrow log_y x + log_x y + log_x x + log_y y = 0 \\ \Rightarrow log_x xy + log_y xy = 0 \\ \Rightarrow \frac{ln ~xy }{ln ~x} = - \frac{ln ~xy }{ln ~y} \\ \Rightarrow (ln ~x + ln ~y)ln ~xy = 0 \\ \mbox{Therefore} \ \ xy = 1$

If we let $y^n=x$ and $x^p=y$ then we are told that $p+n=-2$ . Multiplying these equations yields $(xy)^{p+n}=xy$ . Substituting yields $(xy)^{-2}=xy$ . The only value of $xy$ for which this is satisfied is at $xy=\boxed{1}$ .

Note: This solution is invalid if the problem was defined over complex numbers. Seeing as that is not the case, the solution holds.

Let $log_y x=a$ and $log_x y=b$ . This means that 1) $y^a=x$ and 2) $x^b=y$ . Substituting $y^a$ into the second equation yields $y^{ab}=y^1$ , which shows that ab=1. Then, we define a= $\frac{1}{b}$ . Substituting a into the original a+b equation, we get $\frac{1}{b}$ $+b=-2$ . Multiplying b on both sides, knowing b cannot equal zero, $b^2+2b+1=0$ which gives $b=-1$ . Since that is true, $x^{-1}=y$ , or $\frac{1}{x}=y$ . This shows that $xy=1$