In a Mathematics lesson, Canan performed operations by following the steps below in order. I. $\operatorname { step } \quad : \quad 6 = 1 \cdot 2 \cdot 3 = \mathrm { e } ^ { \ln 1 } \cdot \mathrm { e } ^ { \ln 2 } \cdot \mathrm { e } ^ { \ln 3 }$ II. $\operatorname { step } : \quad e ^ { \ln 1 } \cdot e ^ { \ln 2 } \cdot e ^ { \ln 3 } = e ^ { \ln 1 + \ln 2 + \ln 3 }$ III. step: $\quad e ^ { \ln 1 + \ln 2 + \ln 3 } = e ^ { \ln 6 }$ IV. $\operatorname { step } : : \mathrm { e } ^ { \ln 6 } = \mathrm { e } ^ { \ln ( 2 + 4 ) }$ V. step: $\mathrm { e } ^ { \ln ( 2 + 4 ) } = \mathrm { e } ^ { \ln 2 + \ln 4 }$ VI. $\operatorname { step } : \quad e ^ { \ln 2 + \ln 4 } = e ^ { \ln 2 } \cdot e ^ { \ln 4 }$ VII. step: $e ^ { \ln 2 } \cdot e ^ { \ln 4 } = 2 \cdot 4 = 8$ At the end of these steps, Canan obtained the result $6 = 8$. Accordingly, in which of the numbered steps did Canan make an error? A) II B) III C) IV D) V E) VI
In a Mathematics lesson, Canan performed operations by following the steps below in order.\\
I. $\operatorname { step } \quad : \quad 6 = 1 \cdot 2 \cdot 3 = \mathrm { e } ^ { \ln 1 } \cdot \mathrm { e } ^ { \ln 2 } \cdot \mathrm { e } ^ { \ln 3 }$\\
II. $\operatorname { step } : \quad e ^ { \ln 1 } \cdot e ^ { \ln 2 } \cdot e ^ { \ln 3 } = e ^ { \ln 1 + \ln 2 + \ln 3 }$\\
III. step: $\quad e ^ { \ln 1 + \ln 2 + \ln 3 } = e ^ { \ln 6 }$\\
IV. $\operatorname { step } : : \mathrm { e } ^ { \ln 6 } = \mathrm { e } ^ { \ln ( 2 + 4 ) }$\\
V. step: $\mathrm { e } ^ { \ln ( 2 + 4 ) } = \mathrm { e } ^ { \ln 2 + \ln 4 }$\\
VI. $\operatorname { step } : \quad e ^ { \ln 2 + \ln 4 } = e ^ { \ln 2 } \cdot e ^ { \ln 4 }$\\
VII. step: $e ^ { \ln 2 } \cdot e ^ { \ln 4 } = 2 \cdot 4 = 8$
At the end of these steps, Canan obtained the result $6 = 8$.\\
Accordingly, in which of the numbered steps did Canan make an error?\\
A) II\\
B) III\\
C) IV\\
D) V\\
E) VI