Question
Solve for $x:2^{2x- 1} -9 \times 2^{x - 2} + 1= 0$
✓
Answer
$ 2^{2 x-1}-9 \times 2^{x-2}+1=0$
$2^{2 x} \cdot 2^{-1}-9 x^{2 x} \cdot 2^{-2}+1=0 $
Let $2^x=t$, so $2^{2 x}=t^2$
So, $2^{2 x} \cdot 2^{-1}-9 \times 2^x \cdot 2^{-2}+1=0$
becomes $\frac{t^2}{2}-9 \times \frac{t}{2^2}+1=0$
$ \Rightarrow \frac{t^2}{2}-\frac{9 t}{4}+1=0$
$\Rightarrow 2 t^2-9 t+4=0$
$\Rightarrow 2 t^2-8 t-t+4=0$
$\Rightarrow 2 t(t-4)-1(t-4)=0$
$\Rightarrow( t -4)(2 t -1)=0$
$\Rightarrow t -4=0$ or $2 t -1=0$
$\Rightarrow t =4$ or $t =\frac{1}{2}$
So, $2^x=4$ or $2^x=\frac{1}{2}$
$\Rightarrow 2^x=2^2$ or $2^x=2^{-1}$
$\Rightarrow x =2$ or $x =-1$.
$2^{2 x} \cdot 2^{-1}-9 x^{2 x} \cdot 2^{-2}+1=0 $
Let $2^x=t$, so $2^{2 x}=t^2$
So, $2^{2 x} \cdot 2^{-1}-9 \times 2^x \cdot 2^{-2}+1=0$
becomes $\frac{t^2}{2}-9 \times \frac{t}{2^2}+1=0$
$ \Rightarrow \frac{t^2}{2}-\frac{9 t}{4}+1=0$
$\Rightarrow 2 t^2-9 t+4=0$
$\Rightarrow 2 t^2-8 t-t+4=0$
$\Rightarrow 2 t(t-4)-1(t-4)=0$
$\Rightarrow( t -4)(2 t -1)=0$
$\Rightarrow t -4=0$ or $2 t -1=0$
$\Rightarrow t =4$ or $t =\frac{1}{2}$
So, $2^x=4$ or $2^x=\frac{1}{2}$
$\Rightarrow 2^x=2^2$ or $2^x=2^{-1}$
$\Rightarrow x =2$ or $x =-1$.
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