Solve 9^x-3^x-6=0-n

$$9 ^ { x } - 3 ^ { x } - 6 = 0 - n$$

Solve for n

$$n=3^{x}-9^{x}+6$$

Steps for Solving Linear Equation

Anything plus zero gives itself.

$$9^{x}-3^{x}-6=-n$$

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

$$-n=9^{x}-3^{x}-6$$

Divide both sides by -1.

$$\frac{-n}{-1}=\frac{9^{x}-3^{x}-6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n=\frac{9^{x}-3^{x}-6}{-1}$$

Divide -3^{x}+9^{x}-6 by -1.

$$n=3^{x}-9^{x}+6$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\log_{3}\left(\frac{\sqrt{25-4n}+1}{2}\right)+\frac{2i\pi n_{2}}{\ln(3)}\text{, }n_{2}\in \mathrm{Z}\text{, }&\text{unconditionally}\\x=\log_{3}\left(\frac{-\sqrt{25-4n}+1}{2}\right)+\frac{2i\pi n_{1}}{\ln(3)}\text{, }n_{1}\in \mathrm{Z}\text{, }&n\neq 6\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\log_{3}\left(\frac{\sqrt{25-4n}+1}{2}\right)\text{, }&n\leq \frac{25}{4}\\x=\log_{3}\left(\frac{-\sqrt{25-4n}+1}{2}\right)\text{, }&n>6\text{ and }n\leq \frac{25}{4}\end{matrix}\right.$$