2 log _{10} 5 + log _{10} 8 - frac{1}{2} log | vedclass

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(A) Using the properties of logarithms: $n \log a = \log a^n$ and $\log a + \log b - \log c = \log \left( \frac{a 	imes b}{c} ight)$. Given express

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$2$

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(A) Using the properties of logarithms: $n \log a = \log a^n$ and $\log a + \log b - \log c = \log \left( \frac{a 	imes b}{c} ight)$. Given expression: $2 \log _{10} 5 + \log _{10} 8 - \frac{1}{2} \log _{10} 4$ $= \log _{10} (5^2) + \log _{10} 8 - \log _{10} (4^{1/2})$ $= \log _{10} 25 + \log _{10} 8 - \log _{10} 2$ $= \log _{10} \left( \frac{25 	imes 8}{2} ight)$ $= \log _{10} (100)$ $= \log _{10} (10^2)$ $= 2 \log _{10} 10 = 2 	imes 1 = 2$.

$2 \log _{10} 5 + \log _{10} 8 - \frac{1}{2} \log _{10} 4 = ?$

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