Solve x_{1}x^{2}+x^3-x^4*x^5+6=9^3x+2

$$x _ { 1 } x ^ { 2 } + x ^ { 3 } - x ^ { 4 } \cdot x ^ { 5 } + 6 = 9 ^ { 3 x } + 2$$

Solve for x_1

$$x_{1}=\frac{x^{9}-x^{3}+729^{x}-4}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

To multiply powers of the same base, add their exponents. Add 4 and 5 to get 9.

$$x_{1}x^{2}+x^{3}-x^{9}+6=9^{3x}+2$$

Subtract x^{3} from both sides.

$$x_{1}x^{2}-x^{9}+6=9^{3x}+2-x^{3}$$

Add x^{9} to both sides.

$$x_{1}x^{2}+6=9^{3x}+2-x^{3}+x^{9}$$

Subtract 6 from both sides.

$$x_{1}x^{2}=9^{3x}+2-x^{3}+x^{9}-6$$

Subtract 6 from 2 to get -4.

$$x_{1}x^{2}=9^{3x}-4-x^{3}+x^{9}$$

The equation is in standard form.

$$x^{2}x_{1}=x^{9}-x^{3}+729^{x}-4$$

Divide both sides by x^{2}.

$$\frac{x^{2}x_{1}}{x^{2}}=\frac{x^{9}-x^{3}+729^{x}-4}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$x_{1}=\frac{x^{9}-x^{3}+729^{x}-4}{x^{2}}$$