Question

$2^{2x}-3.2^{(x+2)}+32=0$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {\text{2}}^{2x}-{3.2}^{(x+2)}+32=0 \\ \Rightarrow {\left({\text{2}}^x\right)}^2-{3.2}^x{.2}^2+32=0 \\ {\text{Let 2}}^x\text{be}y. \\ \therefore y^2-12y+32=0 \\ \Rightarrow y^2-8y-4y+32=0 \\ \Rightarrow y(y-8)-4(y-8)=0 \\ \Rightarrow (y-8)=0\text{or}(y-4)=0 \\ \Rightarrow y=8\text{or}y=4 \\ \therefore 2^x=8\mathrm{or}{2}^x=4 \\ \Rightarrow 2^x=2^3\mathrm{or}{2}^x=2^2 \\ \Rightarrow x=2\mathrm{or}3 \\ \text{Hence,}2\text{and 3 are the roots of the given equation}. \end{gathered}$$