Evaluate 2 natural log of e-4 natural log of 1+3 log base 2 of 2+1/2 log base 6 of 1

$2 \ln (e) - 4 \ln (1) + 3 \log_{2} (2) + \frac{1}{2} \log_{6} (1)$

Simplify each term.

The natural logarithm of $e$ is $1$ .

$2 \cdot 1 - 4 \ln (1) + 3 \log_{2} (2) + \frac{1}{2} \cdot \log_{6} (1)$

Multiply $2$ by $1$ .

$2 - 4 \ln (1) + 3 \log_{2} (2) + \frac{1}{2} \cdot \log_{6} (1)$

The natural logarithm of $1$ is $0$ .

$2 - 4 \cdot 0 + 3 \log_{2} (2) + \frac{1}{2} \cdot \log_{6} (1)$

Multiply $- 4$ by $0$ .

$2 + 0 + 3 \log_{2} (2) + \frac{1}{2} \cdot \log_{6} (1)$

Logarithm base $2$ of $2$ is $1$ .

$2 + 0 + 3 \cdot 1 + \frac{1}{2} \cdot \log_{6} (1)$

Multiply $3$ by $1$ .

$2 + 0 + 3 + \frac{1}{2} \cdot \log_{6} (1)$

Logarithm base $6$ of $1$ is $0$ .

$2 + 0 + 3 + \frac{1}{2} \cdot 0$

Multiply $\frac{1}{2}$ by $0$ .

$2 + 0 + 3 + 0$

Simplify by adding numbers.

Add $2$ and $0$ .

$2 + 3 + 0$

Add $2$ and $3$ .

$5 + 0$

Add $5$ and $0$ .

$5$