Solve 2^x-m*2^x+1-m*2^x+2=0

$$2 ^ { x } - m \cdot 2 ^ { x + 1 } - m \cdot 2 ^ { x + 2 } = 0$$

Solve for m

$$m=\frac{1}{6}\approx 0.166666667$$

Steps for Solving Linear Equation

Subtract 2^{x} from both sides. Anything subtracted from zero gives its negation.

$$-m\times 2^{x+1}-m\times 2^{x+2}=-2^{x}$$

Combine all terms containing m.

$$\left(-2^{x+1}-2^{x+2}\right)m=-2^{x}$$

Divide both sides by -2^{x+1}-2^{x+2}.

$$\frac{\left(-2^{x+1}-2^{x+2}\right)m}{-2^{x+1}-2^{x+2}}=-\frac{2^{x}}{-2^{x+1}-2^{x+2}}$$

Dividing by -2^{x+1}-2^{x+2} undoes the multiplication by -2^{x+1}-2^{x+2}.

$$m=-\frac{2^{x}}{-2^{x+1}-2^{x+2}}$$

Divide -2^{x} by -2^{x+1}-2^{x+2}.

$$m=\frac{1}{6}$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

$$m = \frac{1}{6} = 0.16666666666666666$$

Solve for x

$$x\in \mathrm{R}$$

$$m = \frac{1}{6} = 0.16666666666666666$$