Solve the pair of simultaneous equations $\log(x + y) = 0$ ...(1) $\quad\quad2 \log x = \log (y - 1)$ ...(2)

Solve the pair of simultaneous equations 

    $\log(x + y) = 0$             ...(1) 

    $2 \log x = \log (y - 1)$    ...(2) 

SOLUTION: 

$\log (x+y)=0$ in index form becomes $x+y = 10^0$ and so 

                     $x+y=1$      (1)   

In (2) we see that the domain of $x$ and $y$ is restricted by $x>0$ and $y-1$>0 (since the log function has domain positive numbers only)

           so we must have $x>0$ and $y>1$. 

Now using log laws on (2) and converting to index form we have 

                 $\log (x^2) = \log(y-1)$ 

                $\log(x^2) - \log(y-1) = 0$ 

             $\log( x^2/(y-1) )  = 0$

              $x^2/(y-1) = 10^0$             (index form of previous line) 

             $x^2/(y-1) = 1$

            $x^2 = y - 1$

            $y = x^2 + 1$    (2) 

So we now have the two simultaneous equations to solve 

    $x+y=1$             (1)   
    $y = x^2 + 1$     (2) 
(but keeping in mind that $x>0, y>1$ must hold also) 

so,              $x + x^2 +1 = 1$ 
                 
                  $x^2 + x = 0$ 

                 $x ( x + 1 )= 0$

               so this gives $x = 0, -1$ 
             but we must have $x>0$ and both of these solutions fail that condition. 

             So, we conclude that there are NO SOLUTIONS!