The problem asks to find the truth set of two equations. The equations are: (i) $4^{\sqrt{x}} = 8^{2x}$ (ii) $(3^y) \cdot (\frac{1}{27})^y = 81$

1. Problem Description

The problem asks to find the truth set of two equations. The equations are:

(i) $4^{\sqrt{x}} = 8^{2x}$

(ii) $(3^y) \cdot (\frac{1}{27})^y = 81$

2. Solution Steps

(i) Solve $4^{\sqrt{x}} = 8^{2x}$

First, rewrite both sides with the same base,

2. $4 = 2^2$ and $8 = 2^3$.

So the equation becomes:

$(2^2)^{\sqrt{x}} = (2^3)^{2x}$

$2^{2\sqrt{x}} = 2^{6x}$

Since the bases are equal, equate the exponents:

$2\sqrt{x} = 6x$

$\sqrt{x} = 3x$

Square both sides:

$x = 9x^2$

$9x^2 - x = 0$

$x(9x - 1) = 0$

So $x = 0$ or $9x - 1 = 0$, which means $x = \frac{1}{9}$.

Check the solutions:

If $x=0$, $4^{\sqrt{0}} = 4^0 = 1$ and $8^{2(0)} = 8^0 = 1$. So $x=0$ is a solution.

If $x=\frac{1}{9}$, $4^{\sqrt{\frac{1}{9}}} = 4^{\frac{1}{3}} = (2^2)^{\frac{1}{3}} = 2^{\frac{2}{3}}$ and $8^{2(\frac{1}{9})} = 8^{\frac{2}{9}} = (2^3)^{\frac{2}{9}} = 2^{\frac{6}{9}} = 2^{\frac{2}{3}}$. So $x=\frac{1}{9}$ is a solution.

(ii) Solve $(3^y) \cdot (\frac{1}{27})^y = 81$

Rewrite the equation:

$3^y \cdot (27^{-1})^y = 81$

$3^y \cdot (3^{-3})^y = 3^4$

$3^y \cdot 3^{-3y} = 3^4$

$3^{y - 3y} = 3^4$

$3^{-2y} = 3^4$

Since the bases are equal, equate the exponents:

$-2y = 4$

$y = -2$

3. Final Answer

$x = 0, \frac{1}{9}$

$y = -2$