Solve (1/27)^3n+1=frac{sqrt{3}}{81}quadf

$$( \frac { 1 } { 27 } ) ^ { 3 n + 1 } = \frac { \sqrt { 3 } } { 81 } \quad f$$

Solve for f

$$f=\sqrt{3}\times \left(\frac{1}{19683}\right)^{n}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 81.

$$81\times \left(\frac{1}{27}\right)^{3n+1}=\sqrt{3}f$$

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

$$\sqrt{3}f=81\times \left(\frac{1}{27}\right)^{3n+1}$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}f}{\sqrt{3}}=\frac{3}{3^{9n}\sqrt{3}}$$

Dividing by \sqrt{3} undoes the multiplication by \sqrt{3}.

$$f=\frac{3}{3^{9n}\sqrt{3}}$$

Divide \frac{3}{3^{9n}} by \sqrt{3}.

$$f=3^{\frac{1}{2}-9n}$$

Solve for n

$$n=-\frac{\log_{3}\left(f\right)}{9}+\frac{1}{18}$$

$$f>0$$