Without using a calculator, find the value of $x$ such that $\frac{9^{2x+3}}{\sqrt{81^x}} = \left(\frac{1}{3}\right)^{1-5x}$. Solve the equation $\frac{10}{3} \sqrt{3^x} = 3^x + 1$

1. Express the numerator and denominator of the left-hand side in base 3

Numerator: $9^{2x+3} = (3^2)^{2x+3} = 3^{4x+6}$
Denominator: $\sqrt{81^x} = (\sqrt{81})^x = 9^x = (3^2)^x = 3^{2x}$

2. Simplify the left-hand side using the quotient rule for indices

$\text{LHS} = \frac{3^{4x+6}}{3^{2x}} = 3^{4x+6-2x} = 3^{2x+6}$

3. Express the right-hand side in base 3

$\text{RHS} = \left(\frac{1}{3}\right)^{1-5x} = (3^{-1})^{1-5x} = 3^{-1+5x} = 3^{5x-1}$

4. Equate the exponents and solve for $x$

$3^{2x+6} = 3^{5x-1}$
$2x + 6 = 5x - 1$
$3x = 7$
$x = \frac{7}{3}$

1. Use a substitution to transform the exponential equation into a quadratic equation

Let $y = \sqrt{3^x} = 3^{\frac{x}{2}}$. Then $3^x = y^2$.

The equation becomes:
$\frac{10}{3}y = y^2 + 1$
$3y^2 - 10y + 3 = 0$

2. Solve the quadratic equation for $y$

$(3y - 1)(y - 3) = 0$
$y = \frac{1}{3}$ or $y = 3$

3. Substitute back to find the values of $x$

When $y = \frac{1}{3}$:
$3^{\frac{x}{2}} = 3^{-1}$
$\frac{x}{2} = -1 \implies x = -2$

When $y = 3$:
$3^{\frac{x}{2}} = 3^1$
$\frac{x}{2} = 1 \implies x = 2$