Solve the following equations for x:
Question:

Solve the following equations for x:

(i) $2^{2 x}-2^{x+3}+2^{4}=0$

(ii) $3^{2 x+4}+1=2 \times 3^{x+2}$

Solution:

(i) We have, $\Rightarrow 2^{2 x}-2^{x+3}+2^{4}=0$

$\Rightarrow 2^{2 x}+2^{4}=2^{x} \cdot 2^{3}$

$\Rightarrow$ Let $2^{x}=y$

$\Rightarrow y^{2}+2^{4}=y \times 2^{3}$

$\Rightarrow y^{2}-8 y+16=0$

$\Rightarrow y^{2}-4 y-4 y+16=0$

$\Rightarrow y(y-4)-4(y-4)=0$

$\Rightarrow y=4$

$\Rightarrow x^{2}=2^{2}$

$\Rightarrow x=2$

(ii) We have,

$3^{2 x+4}+1=2 \times 3^{x+2}$

$\left(3^{x+2}\right)^{2}+1=2 \times 3^{x+2}$

Let $3^{x+2}=y$

$y^{2}+1=2 y$

$y^{2}-2 y+1=0$

$y^{2}-y-y+1=0$

$y(y-1)-1(y-1)=0$

$(y-1)(y-1)=0$

$y=1$