Solve the following equations: a) 22x+3 - 9 imes 2^x+1 + 1 = 0 b) 3^x imes 3^x - 28 imes 3^x−1 + 3 = 0

$$2^{2x} \cdot 2^3 - 9 \cdot 2^x \cdot 2^1 + 1 = 0$$ $$8 \cdot (2^x)^2 - 18 \cdot 2^x + 1 = 0$$

Substitute $y$ into the equation to form a quadratic equation.

$$8y^2 - 18y + 1 = 0$$

Here, $a=8$, $b=-18$, $c=1$.

$$y = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(8)(1)}}{2(8)}$$ $$y = \frac{18 \pm \sqrt{324 - 32}}{16}$$ $$y = \frac{18 \pm \sqrt{292}}{16}$$ $$y = \frac{18 \pm 2\sqrt{73}}{16}$$ $$y = \frac{9 \pm \sqrt{73}}{8}$$

$$2^x = \frac{9 + \sqrt{73}}{8} \quad \text{or} \quad 2^x = \frac{9 - \sqrt{73}}{8}$$

Take the logarithm base 2 on both sides.

$$x = \log_2\left(\frac{9 + \sqrt{73}}{8}\right) \quad \text{or} \quad x = \log_2\left(\frac{9 - \sqrt{73}}{8}\right)$$

Since $\sqrt{73} \approx 8.54$, $\frac{9 - \sqrt{73}}{8} \approx \frac{9 - 8.54}{8} = \frac{0.46}{8} > 0$. Both solutions are valid.

The solutions are $\boxed{x = \log_2\left(\frac{9 + \sqrt{73}}{8}\right), \log_2\left(\frac{9 - \sqrt{73}}{8}\right)}$.

$$(3^x)^2 - 28 \times \frac{3^x}{3^1} + 3 = 0$$ $$(3^x)^2 - \frac{28}{3} \cdot 3^x + 3 = 0$$

Substitute $y$ into the equation.

$$y^2 - \frac{28}{3}y + 3 = 0$$

Multiply by 3 to clear the fraction:

$$3y^2 - 28y + 9 = 0$$

We look for two numbers that multiply to $3 \times 9 = 27$ and add to $-28$. These numbers are $-27$ and $-1$.

$$3y^2 - 27y - y + 9 = 0$$ $$3y(y - 9) - 1(y - 9) = 0$$ $$(3y - 1)(y - 9) = 0$$

So, $3y - 1 = 0 \implies y = \frac{1}{3}$ or $y - 9 = 0 \implies y = 9$.

For $y = \frac{1}{3}$:

$$3^x = \frac{1}{3}$$ $$3^x = 3^{-1}$$ $$x = -1$$

For $y = 9$:

$$3^x = 9$$ $$3^x = 3^2$$ $$x = 2$$

The solutions are $\boxed{x = -1, 2}$.

$$\frac{2^{2x}}{2^1} - 5 \times \frac{2^x}{2^1} = 12$$ $$\frac{(2^x)^2}{2} - \frac{5 \cdot 2^x}{2} = 12$$

Substitute $y$ into the equation.

$$\frac{y^2}{2} - \frac{5y}{2} = 12$$

Multiply by 2 to clear the fractions:

$$y^2 - 5y = 24$$

We look for two numbers that multiply to $-24$ and add to $-5$. These numbers are $-8$ and 3.

$$(y - 8)(y + 3) = 0$$

So, $y - 8 = 0 \implies y = 8$ or $y + 3 = 0 \implies y = -3$.

For $y = 8$:

$$2^x = 8$$ $$2^x = 2^3$$ $$x = 3$$

For $y = -3$:

$$2^x = -3$$

This equation has no real solution, as $2^x$ is always positive.

The solution is $\boxed{x = 3}$.

$$5^x \cdot 5^3 - \frac{5}{5^x} + 120 = 0$$ $$125 \cdot 5^x - \frac{5}{5^x} + 120 = 0$$

Substitute $y$ into the equation.

$$125y - \frac{5}{y} + 120 = 0$$

Note that $y = 5^x$ must be positive, so $y \neq 0$.

$$125y^2 - 5 + 120y = 0$$

$$125y^2 + 120y - 5 = 0$$

Divide by 5 to simplify:

$$25y^2 + 24y - 1 = 0$$

We look for two numbers that multiply to $25 \times (-1) = -25$ and add to 24. These numbers are 25 and $-1$.

$$25y^2 + 25y - y - 1 = 0$$ $$25y(y + 1) - 1(y + 1) = 0$$ $$(25y - 1)(y + 1) = 0$$

So, $25y - 1 = 0 \implies y = \frac{1}{25}$ or $y + 1 = 0 \implies y = -1$.

For $y = \frac{1}{25}$:

$$5^x = \frac{1}{25}$$ $$5^x = 5^{-2}$$ $$x = -2$$

For $y = -1$:

$$5^x = -1$$

This equation has no real solution, as $5^x$ is always positive.

The solution is $\boxed{x = -2}$.

$$(3^x)^2 - 10 \times \frac{3^x}{3^1} + 1 = 0$$ $$(3^x)^2 - \frac{10}{3} \cdot 3^x + 1 = 0$$

Substitute $y$ into the equation.

$$y^2 - \frac{10}{3}y + 1 = 0$$

Multiply by 3 to clear the fraction:

$$3y^2 - 10y + 3 = 0$$

We look for two numbers that multiply to $3 \times 3 = 9$ and add to $-10$. These numbers are $-9$ and $-1$.

$$3y^2 - 9y - y + 3 = 0$$ $$3y(y - 3) - 1(y - 3) = 0$$ $$(3y - 1)(y - 3) = 0$$

So, $3y - 1 = 0 \implies y = \frac{1}{3}$ or $y - 3 = 0 \implies y = 3$.

For $y = \frac{1}{3}$:

$$3^x = \frac{1}{3}$$ $$3^x = 3^{-1}$$ $$x = -1$$

For $y = 3$:

$$3^x = 3$$ $$3^x = 3^1$$ $$x = 1$$

The solutions are $\boxed{x = -1, 1}$.