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! 

Solve the following equation for x: $$2^{2x + 1} - 9 \times 2^x + 4 = 0$$

Solve the following equation for x: 
    $$2^{2x + 1}  -  9 \times 2^x  +  4  =  0$$

 SOLUTION:  This is a quadratic in $2^x$. 
           $2 (2^x)^2 - 9 (2^x) +4 = 0$ 
         Use a substitution $u=2^x$ and express the equation in terms of $u$.

Therefore:
  $2u^2 - 9u +4 = 0$ 

Factorise to get 
$ ( 2u- 1 )( u - 4 ) = 0 $
$ 2u - 1 = 0$      or     $u - 4 = 0$ 
$ 2u  = 1 $         or     $u = 4$ 
$ u = 1/2$          or     $u=2^2 $

$2^x=2^{-1}$    or     $2^x=2^2$
 

            $\therefore x=-1, 2$