Đáp án đúng: A
Giải chi tiết:Ta có \(A=\log \left( 2017+\log \left( 2016+\log \left( 2015+\log \left( ...+\log \left( 3+\log 2 \right)... \right) \right) \right) \right)\)
\(>\log \left( 2017+\log 2016 \right)>\log \left( 2017+3 \right)=\log 2020\,\,\Rightarrow \,\,A>\log 2020.\)
Áp dụng bất đẳng thức \(\log x < x\,\,\forall x > 1\) ta có
\(\begin{align} 2015+\log \left( 2014+\log \left( \,\,...\,\,+\log \left( 3+\log 2 \right)\,\,...\,\, \right) \right)<2015+2014+\log \left( \,\,...\,\,+\log \left( 3+\log 2 \right)\,\,...\,\, \right) \\ <2015+2014+2013+\,\,...\,\,+3+2=\frac{2017\,\,\times \,\,2014}{2} \end{align}\)
Khi đó \(\log \left( 2016+\log \left( 2015+\log \left( \,\,...\,\,+\log \left( 3+\log 2 \right)\,\,...\,\, \right) \right) \right)<\log \left( 2016+\frac{2017\,\,\times \,\,2014}{2} \right)<4.\)
Vậy \(A<\log \left( 2017+4 \right)=\log 2021\,\,\,\xrightarrow{{}}\,\,\,A\in \left( \log 2020;\,\,\log 2021 \right).\)
Chọn D