Pelo Teorema de Laplace, temos: $d = A_{11}\cdot 10^{x\log 2}$, em que $A_{11}=(-1)^{1+1}\cdot d_{11}=d_{11}$ $d_{11}= \begin{vmatrix} \log {100} & 0\\ 0 & 2^{3x} \end{vmatrix}$ $=$ $2^{3x}\cdot \log {100}$ $=$ $2^{3x+1}$ $\implies$ $A_{11}=2^{3x+1}$ $d = A_{11}\cdot 10^{x\log 2}=A_{11}\cdot 2^{x}$ $\implies$ $d=2^{4x+1}$ Então $\log_2 {d}$ $=$ $\log_2 {2^{4x+1}}$ $=$ $(4x+1)\cdot \log_2 {2}$ $\log_2 {d}$ $=$ $4x+1$ Letra $\mathbb {A}$