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We will prove that for positive number $x$ and $x>1$ , we have $\log_{x}(x+1)>\log_{x+1}(x+2)$ .

Indeed, we have:

$\quad\;\log_{x}(x+1)>\log_{x+1}(x+2)$

$\Leftrightarrow \dfrac{\ln(x+1)}{\ln x}>\dfrac{\ln(x+2)}{\ln(x+1)}$

$\Leftrightarrow \ln^2(x+1)>\ln x*\ln(x+2)$

Applying the Arithmetic Mean Geometric Mean Inequality , we get:

$\ln x+\ln(x+2)\ge2\sqrt{\ln x*\ln(x+2)}$

On the other hand, we have: $(x+1)^2>x(x+2)\Leftrightarrow 2\ln(x+1)>\ln x+\ln(x+2)$

Thus, $\ln(x+1)>\sqrt{\ln x*\ln(x+2)}$ or $\ln^2(x+1)>\ln x*\ln(x+2)$

So the smallest number is $\boxed{E}$ .

The five terms are of the form $\log_n {(n+1)} = \dfrac{\log{(n+1)}}{\log{n}}$ . Let $n = m+1$ , then we have

$\begin{aligned} \log_n {(n+1)} = \dfrac{\log{(m+2)}}{\log{(m+1)}} & \color{#3D99F6}{\propto} \dfrac{\log{(1+\frac{2}{m})}}{\log{(1+\frac{1}{m})}} \quad \color{#3D99F6} {[\propto \text{ means "increases with"}]} \\ & \propto \dfrac{\frac{2}{m} - \frac{1}{2} \left(\frac{2}{m}\right)^2 + \frac{1}{3} \left(\frac{2}{m}\right)^3-...}{\frac{1}{m} - \frac{1}{2} \left(\frac{1}{m}\right)^2 + \frac{1}{3} \left(\frac{1}{m}\right)^3-...} \\ & \propto \frac{2+\frac{2}{m}}{1+\frac{1}{2m}} = \frac{4m+4}{2m+1} = 2 + \frac{2}{2m+1} \\ \Rightarrow \log_n {(n+1)} & \propto \frac{1}{2n-1} \end{aligned}$

This means that the larger the $n$ the smaller the $\log_n {(n+1)}$ . Therefore, the smallest option is $\boxed{(\text{E})} \space \log_{2019} {2020}$

Calculus Method:- 1 2 3 4 5

This is a very intuitive solution. Take these three numbers: $log_2 3$ , $log_3 4$ , and $log_4 5$ . You will notice that the first number $log_2 3$ has to have a larger exponent in order to yield a result of +1. But as the base increases, the exponent power, or the logarithm, needed to have a result 1 greater than base decreases. So that means the logarithm expression decreases with increased bases. Therefore, option E , $log _{2019} 2020$ , would have the smallest value.

Option E A=1.000065212 B=1.000065175 C=1.000065139 D=1.000065102 E=1.000065066

Generalize. Prove log n-1 n > log n n+1 with n in a close neighbour hood of 2015 and 2020 and we see that it has to be the last one E)