Solve 4^x-1/2^x=2^x+1-a

$$\frac{ { 4 }^{ x } -1 }{ { 2 }^{ x } } = { 2 }^{ x+1 } -a$$

Solve for a

$$a=\frac{1}{2^{x}}+2^{x}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2^{x+1}-a=\frac{4^{x}-1}{2^{x}}$$

Subtract 2^{x+1} from both sides.

$$-a=\frac{4^{x}-1}{2^{x}}-2^{x+1}$$

Divide both sides by -1.

$$\frac{-a}{-1}=\frac{-\frac{1}{2^{x}}-2^{x}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a=\frac{-\frac{1}{2^{x}}-2^{x}}{-1}$$

Divide -2^{x}-\frac{1}{2^{x}} by -1.

$$a=\frac{1}{2^{x}}+2^{x}$$

Solve for x

$$x=\log_{2}\left(\frac{\sqrt{a^{2}-4}+a}{2}\right)$$

$$x=\log_{2}\left(\frac{-\sqrt{a^{2}-4}+a}{2}\right)\text{, }a\geq 2$$