Question

Solve the following equations.
$lo{g}^2(4-x)+log(4-x)\cdot log(x+\frac{1}{2})=2lo{g}^2(x+\frac{1}{2})$

Answer 1

$lo{g}^2(4-x)+log(4-x).log(x+4)=2lo{g}^2(x+\frac{1}{2})$

$\Rightarrow log(4-x)\left[log\right(4-x)+log(x+\frac{1}{2}\left)\right]=2lo{g}^2(x+1/2)$

$\Rightarrow log(4-x)log(4-x)(x+1/2)=2lo{g}^2(x+1/2)$

Let $log(4-x)=A;log(x+1/2)=B$

$\Rightarrow A^2+AB=2B^2$

$\Rightarrow A^2+AB-2B^2=0$

$\Rightarrow A^2+2AB-AB-2B^2=0.$

$\Rightarrow A(A+2B)-B(A+2B)=0$

$\Rightarrow (A-B)(A+2B)=0$

$\therefore A=BA=-2B$

$\Rightarrow log(4-x)=log(x+1/2)log(4-x)=log(x+1/2)$

$\Rightarrow 4-x=x+1/2\Rightarrow 4-x=(x+1/2{)}^{-2}$

$\Rightarrow x=7/4\Rightarrow \frac{(2x+1{)}^2}{4}(4-x)=0$

$\Rightarrow (x-4)(2x+1{)}^2=0$

$\therefore x=4$ or $-1/2$

However $x=4$ does not yield any solvable solution as

$log(4-x)$ will be undefined for $x=4$

Similarly $x=-1/2$

$log(x+1/2)$ will be undefined

Hence only solution is $x=7/4$