+26393 exponential problem
4^(x) -2^(x+2) +4=0
\(\begin{array}{|rcll|} \hline 4^{x} -2^{x+2} +4 &=& 0 \\ (2\cdot2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \\ (2^2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \quad & | \quad (2^2)^{x} = 2^{2x} = 2^{x\cdot 2}= (2^x)^2 \\ (2^x)^2 -2^x\cdot2\cdot 2 +2^2 &=& 0 \quad & | \quad \text{binom } \\ (2^x-2)^2 &=& 0 \quad & | \quad \text{square root both sides}\\ 2^x-2 &=& 0 \quad & | \quad +2 \\ 2^x &=& 2 \quad & | \quad 2=2^1 \\ 2^x &=& 2^1 \\ \mathbf{x} &\mathbf{=}& \mathbf{ 1} \\ \hline \end{array}\)
+26393 exponential problem
4^(x) -2^(x+2) +4=0
\(\begin{array}{|rcll|} \hline 4^{x} -2^{x+2} +4 &=& 0 \\ (2\cdot2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \\ (2^2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \quad & | \quad (2^2)^{x} = 2^{2x} = 2^{x\cdot 2}= (2^x)^2 \\ (2^x)^2 -2^x\cdot2\cdot 2 +2^2 &=& 0 \quad & | \quad \text{binom } \\ (2^x-2)^2 &=& 0 \quad & | \quad \text{square root both sides}\\ 2^x-2 &=& 0 \quad & | \quad +2 \\ 2^x &=& 2 \quad & | \quad 2=2^1 \\ 2^x &=& 2^1 \\ \mathbf{x} &\mathbf{=}& \mathbf{ 1} \\ \hline \end{array}\)
