Question

Solve : ${\log }_3(\sqrt{x}+|\sqrt{x}-1|)={\log }_9(4\sqrt{x}-3+4|\sqrt{x}-1|)$

Answer 1

Given

${\log }_3(\sqrt{x}+|\sqrt{x}-1|)={\log }_9(4\sqrt{x}-3+4|\sqrt{x}-1|)\Rightarrow {\log }_3(\sqrt{x}+|\sqrt{x}-1|)={\log }_{3^2}(4\sqrt{x}-3+4|\sqrt{x}-1|)\Rightarrow {\log }_3(\sqrt{x}+|\sqrt{x}-1|)=\frac{1}{2}{\log }_3(4\sqrt{x}-3+4|\sqrt{x}-1|)({\log }_{a^m}b=\frac{1}{m}{\log }_{a}b)\Rightarrow \sqrt{x}+|\sqrt{x}-1|=\sqrt{(4\sqrt{x}-3+4|\sqrt{x}-1|)}$

Squaring both sides we get ,

$\Rightarrow x+x+1-2\sqrt{x}+2\sqrt{x}|\sqrt{x}-1|=4\sqrt{x}-3+4|\sqrt{x}-1|-\left(i\right)$

Case1:$\sqrt{x}-1\ge 0\Rightarrow x\ge 1$

Equation $\left(i\right)$ then reduces to

$2x+1-2\sqrt{x}+2x-2\sqrt{x}=4\sqrt{x}-3+4\sqrt{x}-4\Rightarrow 4x-4\sqrt{x}+1=8\sqrt{x}-7\Rightarrow 4x-12\sqrt{x}+8=0\Rightarrow x-3\sqrt{x}+2=0\Rightarrow x-2\sqrt{x}-\sqrt{x}+2=0\Rightarrow \sqrt{x}(\sqrt{x}-2)-1(\sqrt{x}-2)=0\Rightarrow (\sqrt{x}-1)(\sqrt{x}-2)=0\therefore x=1orx=4$

In the region $x\ge 1$,both $x=1$ & $4$ are acceptable.

Case2:$\sqrt{x}-1<0\Rightarrow x<1$ but $x>0$ i.e.,$xϵ(0,1)$

Equation $\left(i\right)$ then reduces to

$2x+1-2\sqrt{x}+2\sqrt{x}-2x=4\sqrt{x}-3+4\sqrt{x}1=1$

which is always true.

$\therefore$ the given equality

${\log }_3(\sqrt{x}+|\sqrt{x}-1|)={\log }_9(4\sqrt{x}-3+4|\sqrt{x}-1|)$

holds true for $xϵ(0,1)\cup \{1,4\}$