219 2) 2ˣ⁻³ = 3³⁻ˣ;

2) 2ˣ⁻³ = 3³⁻ˣ

$$2^{x-3} = 3^{3-x}$$

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

$$2^{x-3} = \frac{27}{3^x}$$

Прологарифмируем обе части:

$$(x-3)\log(2) = \log(27) - x\log(3)$$

$$x\log(2) - 3\log(2) = \log(27) - x\log(3)$$

$$x\log(2) + x\log(3) = \log(27) + 3\log(2)$$

$$x(\log(2) + \log(3)) = \log(27) + \log(8)$$

$$x = \frac{\log(27 \cdot 8)}{\log(2 \cdot 3)} = \frac{\log(216)}{\log(6)}$$

$$x = \frac{\log(6^3)}{\log(6)} = \frac{3 \log(6)}{\log(6)} = 3$$

Ответ: 3