Find the value of $\dfrac{1}{3}log_{10}125−2log_{10}4+log_{10}32$
= $\dfrac{1}{3}log_{10}125−2log_{10}4+log_{10}32$
=$log_{10}(125^{1/3})−log_{10}(4^2)+log_{10}32$
=$log_{10}5−log_{10}16+log_{10}32$
=$log_{10}\left(\dfrac{5\times32}{16}\right)$
=$log_{10}10$
=1
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