Respuesta :

LammettHash

32 = 2⁵ and 8 = 2³, so you can rewrite the equation as

[tex]32^{2c}=8^{c+7}\implies(2^5)^{2c}=(2^3)^{c+7}\implies 2^{10c}=2^{3c+21}[/tex]

The bases of both sides are equal, so the exponents must be too. Solve for c :

10c = 3c + 21

7c = 21

c = 3