Solve sqrt2^x=sqrt{2}quada=?

$$\sqrt { 2 } ^ { x } = \sqrt { 2 } \quad a = ?$$

Solve for a

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

Steps for Solving Linear Equation

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

$$\sqrt{2}a=\left(\sqrt{2}\right)^{x}$$

Divide both sides by \sqrt{2}.

$$\frac{\sqrt{2}a}{\sqrt{2}}=\frac{2^{\frac{x}{2}}}{\sqrt{2}}$$

Dividing by \sqrt{2} undoes the multiplication by \sqrt{2}.

$$a=\frac{2^{\frac{x}{2}}}{\sqrt{2}}$$

Divide 2^{\frac{x}{2}} by \sqrt{2}.

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

Solve for x (complex solution)

$$x=\frac{2\ln(a)+\ln(2)}{\ln(2)}+\frac{4\pi n_{1}i}{\ln(2)}$$

$$n_{1}\in \mathrm{Z}$$

$$a\neq 0$$

Solve for x

$$x=2\log_{2}\left(a\right)+1$$

$$a>0$$