Solve for x (1/27)^(x^2)=9^(3x+8)*(1/81)^(x^2)

${(\frac{1}{27})}^{x^{2}} = 9^{3 x + 8} \cdot {(\frac{1}{81})}^{x^{2}}$

Simplify ${(\frac{1}{27})}^{x^{2}}$ .

Apply the product rule to $\frac{1}{27}$ .

$\frac{1^{x^{2}}}{27^{x^{2}}} = 9^{3 x + 8} \cdot {(\frac{1}{81})}^{x^{2}}$

One to any power is one.

$\frac{1}{27^{x^{2}}} = 9^{3 x + 8} \cdot {(\frac{1}{81})}^{x^{2}}$

Simplify $9^{3 x + 8} \cdot {(\frac{1}{81})}^{x^{2}}$ .

Simplify the expression.

Apply the product rule to $\frac{1}{81}$ .

$\frac{1}{27^{x^{2}}} = 9^{3 x + 8} \cdot \frac{1^{x^{2}}}{81^{x^{2}}}$

One to any power is one.

$\frac{1}{27^{x^{2}}} = 9^{3 x + 8} \cdot \frac{1}{81^{x^{2}}}$

Combine $9^{3 x + 8}$ and $\frac{1}{81^{x^{2}}}$ .

$\frac{1}{27^{x^{2}}} = \frac{9^{3 x + 8}}{81^{x^{2}}}$

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx - 2, 8, 8.1, 15$