Let me extract the problems from the image and solve them for you. Here are the equations from the image:

Problems:

1. $8^x - 4^x = 2^{x+1}$
2. $4 \sqrt{3x^2 - 2x + 1} + 2 = 9 \cdot 2\sqrt{3x^2 - 2x}$
3. $4 \cdot 3^{x^2} + x - 8 = 2 \cdot \frac{x^2 + x}{3}$
4. $4 \cdot 2^{2x} - 6 = 18 \cdot 3^{2x}$
5. $5 \cdot 2^{3x} - 3 \cdot 2^{5-3x} = -7$
6. $x = (\sqrt{x})^x$

I'll solve them step by step.

1. $8^x - 4^x = 2^{x+1}$

Rewrite terms in base $2$:

• $8^x = (2^3)^x = 2^{3x}$
• $4^x = (2^2)^x = 2^{2x}$

Substitute into the equation: $2^{3x} - 2^{2x} = 2^{x+1}$

Factor out $2^{x}$:

$2^{x} (2^{2x} - 2^x - 2) = 0$

For $2^x \neq 0$, solve: $2^{2x} - 2^x - 2 = 0$

Let $y = 2^x$. Then: $y^2 - y - 2 = 0$

Factorize: $(y - 2)(y + 1) = 0$

Solutions: $y = 2 \quad \text{or} \quad y = -1$

Since $y = 2^x > 0$, $y = 2$. Therefore: $2^x = 2 \implies x = 1$

2. $4 \sqrt{3x^2 - 2x + 1} + 2 = 9 \cdot 2\sqrt{3x^2 - 2x}$

Simplify:

Let $y = \sqrt{3x^2 - 2x}$. Then: $4y + 2 = 18y$

Solve for $y$: $2 = 18y - 4y \implies 2 = 14y \implies y = \frac{1}{7}$

Substitute back: $\sqrt{3x^2 - 2x} = \frac{1}{7}$

Square both sides: $3x^2 - 2x = \frac{1}{49}$

Multiply through by 49: $147x^2 - 98x - 1 = 0$

Solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 147$, $b = -98$, $c = -1$: $x = \frac{-(-98) \pm \sqrt{(-98)^2 - 4(147)(-1)}}{2(147)}$

Calculate: $x = \frac{98 \pm \sqrt{9604 + 588}}{294} = \frac{98 \pm \sqrt{10192}}{294}$

Simplify: $x = \frac{98 \pm 101.96}{294}$

Two solutions: $x = \frac{199.96}{294} \approx 0.68 \quad \text{or} \quad x = \frac{-3.96}{294} \approx -0.013$

3. $4 \cdot 3^{x^2} + x - 8 = 2 \cdot \frac{x^2 + x}{3}$

Rewriting: $4 \cdot 3^{x^2} + x - 8 = \frac{2x^2 + 2x}{3}$

Multiply through by 3: $12 \cdot 3^{x^2} + 3x - 24 = 2x^2 + 2x$

Rearrange: $12 \cdot 3^{x^2} - 2x^2 + x - 24 = 0$

This is a non-linear equation and can be solved numerically or through iterative methods.

4. $4 \cdot 2^{2x} - 6 = 18 \cdot 3^{2x}$

Rewrite terms:

$4 \cdot (2^x)^2 - 6 = 18 \cdot (3^x)^2$

Let $y = 2^x$ and $z = 3^x$. Then $z = \frac{y^3}{2^3}$: ...

Would you like me to continue solving all the problems or focus on specific ones in more depth? Also, feel free to clarify any doubts.

Related Questions:

1. How is the quadratic formula derived, and how can it be applied?
2. What are efficient ways to solve equations with exponents?
3. Why is $y = 2^x$ substitution used in exponential equations?
4. How can numerical methods help solve higher-degree non-linear equations?
5. What tools can simplify solving equations involving square roots?

Tip:

For equations involving exponents, always look for opportunities to express all terms in the same base—it simplifies solving considerably.