Question
Solve for $x:2^{2x+3} - 9 x 2^x+ 1 = 0$
✓
Answer
$2^{2 x+3}-9 \times 2^x+1=0$
$2^{2 x} \cdot 2^3-9 \times 2^x+1=0$
Put $2^x=t$, so, $2^{2 x}=t^2$
So, $2^{2 x} \cdot 2^3-9 \times 2^x+1=0$
becomes $8 t^2-9 t+1=0$
$\Rightarrow 8 t^2-8 t-t+1=0$
$\Rightarrow 8 t ( t -1)-( t -1)=0$
$\Rightarrow t -1=0$ or $8 t -1=0$
$\Rightarrow t =1$ or $t =\frac{1}{8}$
$\Rightarrow 2^{ x }=1$ or $2^{ x }=\frac{1}{2^3}$
$\Rightarrow 2^{ x }=2^0$ or $2^{ x }=2^{-3}$
$\Rightarrow x =0$ or $x =-3$
$2^{2 x} \cdot 2^3-9 \times 2^x+1=0$
Put $2^x=t$, so, $2^{2 x}=t^2$
So, $2^{2 x} \cdot 2^3-9 \times 2^x+1=0$
becomes $8 t^2-9 t+1=0$
$\Rightarrow 8 t^2-8 t-t+1=0$
$\Rightarrow 8 t ( t -1)-( t -1)=0$
$\Rightarrow t -1=0$ or $8 t -1=0$
$\Rightarrow t =1$ or $t =\frac{1}{8}$
$\Rightarrow 2^{ x }=1$ or $2^{ x }=\frac{1}{2^3}$
$\Rightarrow 2^{ x }=2^0$ or $2^{ x }=2^{-3}$
$\Rightarrow x =0$ or $x =-3$
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