Question

Which is smaller?: 2 or $(log_{e^{-1}}2 + log_{2}e^{-1})$

Answer 1

$(log_{e^{-1}}2 + log_{2}e^{-1})$

Answer 2

Let the given expression be $E = log_{e^{-1}}2 + log_{2}e^{-1}$. Using logarithm properties, $E = -log_e 2 - log_2 e$. Let $x = log_e 2$. Then $log_2 e = \frac{1}{x}$. So, $E = -(x + \frac{1}{x})$. By AM-GM inequality, for $x>0$, $x + \frac{1}{x} \ge 2$. Since $x = log_e 2 \ne 1$, $x + \frac{1}{x} > 2$. Therefore, $E = -(x + \frac{1}{x}) < -2$. Since $E < -2$, $E$ is smaller than $2$.