The image contains two mathematical problems written in Ukrainian. They are asking to solve equations:

1. $( \frac{1}{2} )^{x-2} = ( \frac{2}{3} )^{2x-4}$
2. $4^x - 3 \cdot 2^x - 4 = 0$

Let's solve them one by one:

Problem 1:

$( \frac{1}{2} )^{x-2} = ( \frac{2}{3} )^{2x-4}$ Take the logarithm of both sides, or rewrite the bases to solve for $x$.

Problem 2:

$4^x - 3 \cdot 2^x - 4 = 0$ Notice that $4^x = (2^2)^x = (2^x)^2$, so let $y = 2^x$, reducing the equation to a quadratic form for $y$.

Let me solve these equations step by step.### Solution for Problem 1: The equation $( \frac{1}{2} )^{x-2} = ( \frac{2}{3} )^{2x-4}$ has the solution: $x = 2$

Solution for Problem 2:

The equation $4^x - 3 \cdot 2^x - 4 = 0$ reduces to a quadratic form, and its solutions for $x$ are: $x = 2 \quad \text{or} \quad x = \frac{i\pi}{\log 2} \quad (\text{complex solution})$ The real solution is $x = 2$.

Would you like further details or clarifications on these solutions?

Related Questions:

1. How do logarithms help in solving exponential equations?
2. What other methods could be used to solve equations with different bases?
3. How do you solve quadratic equations in general?
4. What is the significance of complex solutions in real-world problems?
5. Can every exponential equation be solved by logarithms?

Tip: When dealing with exponential equations, always check if you can express terms with the same base to simplify the solving process.