Solve 3^x=kquad3^2x+1=

$$3 ^ { x } = k \quad 3 ^ { 2 x + 1 } =$$

Solve for k

$$k=\frac{1}{3\times 3^{x}}$$

Steps for Solving Linear Equation

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

$$k\times 3^{2x+1}=3^{x}$$

The equation is in standard form.

$$3^{2x+1}k=3^{x}$$

Divide both sides by 3^{2x+1}.

$$\frac{3^{2x+1}k}{3^{2x+1}}=\frac{3^{x}}{3^{2x+1}}$$

Dividing by 3^{2x+1} undoes the multiplication by 3^{2x+1}.

$$k=\frac{3^{x}}{3^{2x+1}}$$

Divide 3^{x} by 3^{2x+1}.

$$k=3^{-x-1}$$

Solve for x (complex solution)

$$x=-\log_{3}\left(k\right)+\frac{2\pi n_{1}i}{\ln(3)}-1$$

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

$$k\neq 0$$

Solve for x

$$x=-\log_{3}\left(k\right)-1$$

$$k>0$$