Question
Solve : $8 \times 2^{2x} + 4 \times 2^{x + 1} = 1 + 2^x$
✓
Answer
$8 \times 2^{2 x}+4 \times 2^{x+1}=1+2^x$
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1=1+2^x$
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1-1-2^x=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+2^x \times(8-1)-1=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+7\left(2^x\right)-1=0$
$\Rightarrow 8 y^2+7 y-1=0$
$\left[y=2^x\right]$
$ \Rightarrow 8 y^2+8 y-y-1=0$
$ \Rightarrow 8 y(y+1)-1(y+1)=0$
$ \Rightarrow(8 y-1)(y+1)=0$
$ \Rightarrow 8 y=1$ or $y=-1$
$ \Rightarrow y=\frac{1}{8}$ or $y=-1$
$ \Rightarrow 2^x=\frac{1}{8}$ or $2^x=-1$
$ \Rightarrow 2^x=\frac{1}{2^3}$ or $2^x=-1$
$ \Rightarrow 2^x=2^{-3}$ or $2^x=-1$
$ \Rightarrow x=-3$
$[\because 2^x=-1]$ is not possible.
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1=1+2^x$
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1-1-2^x=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+2^x \times(8-1)-1=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+7\left(2^x\right)-1=0$
$\Rightarrow 8 y^2+7 y-1=0$
$\left[y=2^x\right]$
$ \Rightarrow 8 y^2+8 y-y-1=0$
$ \Rightarrow 8 y(y+1)-1(y+1)=0$
$ \Rightarrow(8 y-1)(y+1)=0$
$ \Rightarrow 8 y=1$ or $y=-1$
$ \Rightarrow y=\frac{1}{8}$ or $y=-1$
$ \Rightarrow 2^x=\frac{1}{8}$ or $2^x=-1$
$ \Rightarrow 2^x=\frac{1}{2^3}$ or $2^x=-1$
$ \Rightarrow 2^x=2^{-3}$ or $2^x=-1$
$ \Rightarrow x=-3$
$[\because 2^x=-1]$ is not possible.
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