Question
Solve for x : $9^{x+2}-6.3^{x+1}+1=0$
✓
Answer
Given equation $9^{x+2}-6.3^{x+1}+1=0$
$\Rightarrow 9^x \cdot 9^2-6 \cdot 3^x \cdot 3^1+1=0$
$\Rightarrow 81 \cdot\left(3^2\right)^x-18 \cdot 3^x+1=0$
$\Rightarrow 81 \cdot 3^{2 x}-18 \cdot 3^x+1=0$
Putting $3^x=y$, then it becomes $81 y^2-18 y+1=0$
$\Rightarrow 81 y^2-9 y-9 y+1=0$
$\Rightarrow 9 y(9 y-1)-1(9 y-1)=0$
$\Rightarrow(9 y-1)(9 y-1)=0$
$\Rightarrow 9 y=1$
$\Rightarrow y =\frac{1}{9}$
$\text { But } 3^{ x }=\frac{1}{9}=\frac{1}{3^2}=3^{-2}$
$\therefore x =-2$
Hence, the required root is $-2.$
$\Rightarrow 9^x \cdot 9^2-6 \cdot 3^x \cdot 3^1+1=0$
$\Rightarrow 81 \cdot\left(3^2\right)^x-18 \cdot 3^x+1=0$
$\Rightarrow 81 \cdot 3^{2 x}-18 \cdot 3^x+1=0$
Putting $3^x=y$, then it becomes $81 y^2-18 y+1=0$
$\Rightarrow 81 y^2-9 y-9 y+1=0$
$\Rightarrow 9 y(9 y-1)-1(9 y-1)=0$
$\Rightarrow(9 y-1)(9 y-1)=0$
$\Rightarrow 9 y=1$
$\Rightarrow y =\frac{1}{9}$
$\text { But } 3^{ x }=\frac{1}{9}=\frac{1}{3^2}=3^{-2}$
$\therefore x =-2$
Hence, the required root is $-2.$
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