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# exponential problem

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4^(x) -2^(x+2) +4=0

Guest Oct 26, 2017

#1
+18715
+2

exponential problem

4^(x) -2^(x+2) +4=0

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

heureka  Oct 26, 2017
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#1
+18715
+2

exponential problem

4^(x) -2^(x+2) +4=0

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

heureka  Oct 26, 2017

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