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\[3(\log_{3}14-\log_{3}7)+ 2 \log_{3}5\]

\[\log(14)-\log(7)=\log(\frac{14}{7})=\log(2)\] is a start

then
\[3\log(2)=\log(2^3)=\log(8)\]

and finally
\[\log(8)=2\log(5)\log(8)+\log(5^2)=\log(8\times 25)=\log(200)\]