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The value of $\left(\frac{1}{3}log_{10}125-2log_{10}4+log_{10}32\right)$ is :

Answer:

Explanation:

$\frac{1}{3}log_{10}125$ $-2log_{10}4$ $+log_{10}32$

$=log_{10}(125)^{1/3}$ $-log_{10}(4)^{2}$ $+log_{10}32$

$=log_{10}5$ $-log_{10}16$ $+log_{10}32$

$=log_{10}\left(\frac{5\times 32}{16}\right)$ $=log_{10} 10$ = 1.