Let's go through each question and solve them step-by-step.

1. Solve for $x$ in:

$2(4^x) - 10(2^x) + 8 = 0$

Solution: Let $y = 2^x$. Then $4^x = (2^2)^x = (2^x)^2 = y^2$. Substituting, the equation becomes: $2y^2 - 10y + 8 = 0$ Simplify by dividing by 2: $y^2 - 5y + 4 = 0$ Factor the quadratic: $(y - 4)(y - 1) = 0$ So, $y = 4$ or $y = 1$.

Now, since $y = 2^x$:

• If $2^x = 4$, then $x = 2$.
• If $2^x = 1$, then $x = 0$.

Answer: $x = 2$ or $x = 0$.

2. Solve for $x$ in:

$2^{2x} - 18 \cdot 2^x = 40$

Solution: Let $y = 2^x$. Then $2^{2x} = (2^x)^2 = y^2$. Substitute to get: $y^2 - 18y - 40 = 0$ Solve this quadratic equation using the quadratic formula: $y = \frac{18 \pm \sqrt{(-18)^2 + 4 \cdot 40}}{2}$ (Continue solving for values of $y$ and then $x$)

3. Solve for $x$ given:

$\left(\frac{1}{8}\right)^x \times 64^2 = 256$

4. Simplify:

$\frac{3^{n+3} - 3 \times 3^n}{4 \times 3^{n+2}}$

5. Solve for $x$ in the equation:

$64^x - 121 = 17$

6. Find $x$ if:

$3^{2x+3} + 1 = 28$

Would you like details on any of the remaining steps or solutions?