We are given three exponential equations to solve: (ii) $2^{2y+1} - 15(2^y) = 8$ (iii) $2^{x+3} - 15 = 2^{1-x}$ (v) $3^{2x} - 3^{x+2} = 3^{1+x} - 27$

1. Problem Description

We are given three exponential equations to solve:

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

(iii) $2^{x+3} - 15 = 2^{1-x}$

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

2. Solution Steps

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

We can rewrite the equation as:

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

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

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

We can factor this quadratic equation:

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

So, $2z+1 = 0$ or $z-8 = 0$.

If $2z+1 = 0$, then $z = -\frac{1}{2}$. Since $z = 2^y$, we have $2^y = -\frac{1}{2}$. However, $2^y$ must be positive, so this case has no solution.

If $z-8 = 0$, then $z = 8$. So, $2^y = 8 = 2^3$. Therefore, $y = 3$.

(iii) $2^{x+3} - 15 = 2^{1-x}$

$2^x \cdot 2^3 - 15 = 2^1 \cdot 2^{-x}$

$8 \cdot 2^x - 15 = \frac{2}{2^x}$

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

$8z - 15 = \frac{2}{z}$

Multiplying by $z$, we get:

$8z^2 - 15z = 2$

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

We can factor this quadratic equation:

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

So, $8z+1 = 0$ or $z-2 = 0$.

If $8z+1 = 0$, then $z = -\frac{1}{8}$. Since $z = 2^x$, we have $2^x = -\frac{1}{8}$. However, $2^x$ must be positive, so this case has no solution.

If $z-2 = 0$, then $z = 2$. So, $2^x = 2 = 2^1$. Therefore, $x = 1$.

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

$(3^x)^2 - 3^x \cdot 3^2 = 3 \cdot 3^x - 27$

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

Let $z = 3^x$. Then the equation becomes:

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

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

We can factor this quadratic equation:

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

So, $z-3 = 0$ or $z-9 = 0$.

If $z-3 = 0$, then $z = 3$. So, $3^x = 3 = 3^1$. Therefore, $x = 1$.

If $z-9 = 0$, then $z = 9$. So, $3^x = 9 = 3^2$. Therefore, $x = 2$.

Thus, $x = 1$ or $x = 2$.

3. Final Answer

$y = 3$

$x = 1$

$x = 1, 2$