Andika nambari zote katika equation kama nguvu za base ya kawaida, ambayo ni 2.

Tunayo $4 = 2^2$, $8 = 2^3$, na $32 = 2^5$.
Hivyo, equation inakuwa:

$$(2^2)^{3x} \times 2^3 = (2^{-5})^{2x-3}$$

Tumia sheria ya exponents $(a^m)^n = a^{mn}$:

$$2^{2 \times 3x} \times 2^3 = 2^{-5(2x-3)}$$ $$2^{6x} \times 2^3 = 2^{-10x+15}$$

Ongeza $10x$ pande zote mbili:

$$6x + 10x + 3 = 15$$ $$16x + 3 = 15$$

Toa 3 pande zote mbili:

$$16x = 15 - 3$$ $$16x = 12$$

Gawanya kwa 16:

$$x = \frac{12}{16}$$

Rahisisha sehemu:

$$x = \frac{3}{4}$$

Jibu la mwisho ni $\frac{3}{4}$.