Solve the following equation:

$$
4^{x}-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1}
$$
Let's transform the equation and perform the following transformations:

$$
4^{x}+2^{2 x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}
$$

which is

$$
2^{2 x-1}(2+1)=3^{x-\frac{1}{2}}(3+1)
$$

Dividing both sides by \(3 \cdot 2^{2}\)

$$
2^{2 x-3}=3^{x-\frac{3}{2}}, \quad \text { or } \quad 4^{x-\frac{3}{2}}=3^{x-\frac{3}{2}}
$$

Powers with different bases can only be equal if the exponent is 0, which in our case means that

$$
x=\frac{3}{2}
$$

József Kovács (Pécs, Bányaip. t. II. o. t.)