The problem asks us to solve for $x$ or $y$ in the given equations. The equations are: (i) $2^{2x} - 5(2^x) + 4 = 0$ (ii) $2^{2y+1} - 15(2^y) = 8$ (iv) $2^{3x+1} - 3(2^{2x}) + 2^{x+1} = 2^x$ (v) $3^{2x} - 3^{x+2} = 3^{1+x} - 27$

1. Problem Description

The problem asks us to solve for $x$ or $y$ in the given equations. The equations are:

(i) $2^{2x} - 5(2^x) + 4 = 0$

(ii) $2^{2y+1} - 15(2^y) = 8$

(iv) $2^{3x+1} - 3(2^{2x}) + 2^{x+1} = 2^x$

(v) $3^{2x} - 3^{x+2} = 3^{1+x} - 27$

2. Solution Steps

(i) $2^{2x} - 5(2^x) + 4 = 0$

Let $u = 2^x$. Then the equation becomes:

$u^2 - 5u + 4 = 0$

$(u - 4)(u - 1) = 0$

$u = 4$ or $u = 1$

$2^x = 4 = 2^2$ or $2^x = 1 = 2^0$

$x = 2$ or $x = 0$

(ii) $2^{2y+1} - 15(2^y) = 8$

$2^{2y} \cdot 2^1 - 15(2^y) = 8$

$2(2^y)^2 - 15(2^y) - 8 = 0$

Let $v = 2^y$. Then the equation becomes:

$2v^2 - 15v - 8 = 0$

$2v^2 - 16v + v - 8 = 0$

$2v(v - 8) + 1(v - 8) = 0$

$(2v + 1)(v - 8) = 0$

$v = 8$ or $v = -\frac{1}{2}$

Since $v = 2^y$, $2^y = 8 = 2^3$ or $2^y = -\frac{1}{2}$. $2^y$ can never be negative, so we only have

$y = 3$

(iv) $2^{3x+1} - 3(2^{2x}) + 2^{x+1} = 2^x$

$2^{3x} \cdot 2^1 - 3(2^{2x}) + 2^x \cdot 2^1 = 2^x$

$2(2^x)^3 - 3(2^x)^2 + 2(2^x) = 2^x$

$2(2^x)^3 - 3(2^x)^2 + 2(2^x) - 2^x = 0$

$2(2^x)^3 - 3(2^x)^2 + (2^x) = 0$

$2^x[2(2^x)^2 - 3(2^x) + 1] = 0$

Since $2^x$ is never zero, we have

$2(2^x)^2 - 3(2^x) + 1 = 0$

Let $w = 2^x$.

$2w^2 - 3w + 1 = 0$

$2w^2 - 2w - w + 1 = 0$

$2w(w-1) - (w-1) = 0$

$(2w-1)(w-1) = 0$

$w = 1$ or $w = \frac{1}{2}$

$2^x = 1 = 2^0$ or $2^x = \frac{1}{2} = 2^{-1}$

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

(v) $3^{2x} - 3^{x+2} = 3^{1+x} - 27$

$3^{2x} - 3^x \cdot 3^2 = 3 \cdot 3^x - 27$

$(3^x)^2 - 9(3^x) = 3(3^x) - 27$

$(3^x)^2 - 12(3^x) + 27 = 0$

Let $z = 3^x$

$z^2 - 12z + 27 = 0$

$z^2 - 9z - 3z + 27 = 0$

$z(z-9) - 3(z-9) = 0$

$(z-3)(z-9) = 0$

$z = 3$ or $z = 9$

$3^x = 3 = 3^1$ or $3^x = 9 = 3^2$

$x = 1$ or $x = 2$

3. Final Answer

(i) $x = 0, 2$

(ii) $y = 3$

(iv) $x = -1, 0$

(v) $x = 1, 2$