find the value.\n$lnleft(\frac{1}{e^{12}}\right)$\n$lnleft(\frac{1}{e^{12}}\right)=square$

Explanation:

Step1: Use logarithm property

We know that $\ln(\frac{a}{b})=\ln(a)-\ln(b)$ and $\ln(e^x) = x$. So, $\ln(\frac{1}{e^{12}})=\ln(1)-\ln(e^{12})$.

Step2: Evaluate $\ln(1)$ and $\ln(e^{12})$

Since $\ln(1) = 0$ and $\ln(e^{12})=12$, then $\ln(1)-\ln(e^{12})=0 - 12$.

Answer:

$- 12$