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

1. Problem Description

We are asked to find the truth set of two equations:

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

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

2. Solution Steps

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

Express both sides with the same base, 2:

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

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

Since the bases are equal, we can equate the exponents:

$2\sqrt{x} = 6x$

$\sqrt{x} = 3x$

Square both sides:

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

$x = 9x^2$

$9x^2 - x = 0$

$x(9x - 1) = 0$

$x = 0$ or $9x - 1 = 0$

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

Check $x=0$:

$4^{\sqrt{0}} = 4^0 = 1$

$8^{2(0)} = 8^0 = 1$

So $x=0$ is a solution.

Check $x=\frac{1}{9}$:

$4^{\sqrt{\frac{1}{9}}} = 4^{\frac{1}{3}} = (2^2)^{\frac{1}{3}} = 2^{\frac{2}{3}}$

$8^{2(\frac{1}{9})} = 8^{\frac{2}{9}} = (2^3)^{\frac{2}{9}} = 2^{\frac{6}{9}} = 2^{\frac{2}{3}}$

So $x=\frac{1}{9}$ is a solution.

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

$3^{y^2} \cdot (3^{-3})^y = 3^4$

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

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

Since the bases are equal, we can equate the exponents:

$y^2 - 3y = 4$

$y^2 - 3y - 4 = 0$

$(y - 4)(y + 1) = 0$

$y - 4 = 0$ or $y + 1 = 0$

$y = 4$ or $y = -1$

3. Final Answer

(i) $x = 0, \frac{1}{9}$

(ii) $y = 4, -1$