Question

Solve the equation
${\log }_{x^2+6x+8}{\log }_{{2x}^2+2x+3}(x^2-2x)=0$
find the square of the solution

Answer 1

Given equation
${\log }_{x^2+6x+8}{\log }_{{2x}^2+2x+3}(x^2-2x)=0$
$\Rightarrow {\log }_{{2x}^2+2x+3}(x^2-2x)={(x^2+6x+8)}^0$
$\Rightarrow {\log }_{{2x}^2+2x+3}(x^2-2x)=1$
$\Rightarrow x^2-2x={({2x}^2+2x+3)}^1$
$\Rightarrow x^2+4x+3=0$
$\therefore x=-1,-3$
From the given equation , it follows that
$x^2+6x+8>0$ , ${2x}^2+2x+3>0$ and $x^2-2x>0$
$\Rightarrow (x+4)(x+2)>0$ , ${(x+\frac{1}{2})}^2+\frac{5}{4}>0$ and $x(x-2)>0$
$\Rightarrow x\in (-\infty ,-4)\cup (-2,\infty )$ ......(1)
$x\in (-\infty ,0)\cup (2,\infty )..............\left(2\right)$
But $x=-3$ does satisfies (1) and (2)
$\therefore x=-3$ is not solution of the given equation.
But x=-1 satisfied both conditions (1) and (2)
$\therefore x=-1$ is only solution of the given equation.
Square of the solution is $(-1{)}^2=1$