Question
Evaluate the following without using tables :$\log_{10}8 + \log_{10}25 + 2 \log_{10}3 - \log_{10}18$
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Answer
Consider the given expression
$\log _{10} 8+\log _{10} 25+2 \log _{10} 3-\log _{10} 18$
$=\log _{10} 8+\log _{10} 25+\log _{10} 3^2-\log _{10} 18 \ldots .\left[n \log _a m=\log _a m^n\right]$
$=\log _{10} 8+\log _{10} 25+\log _{10} 9-\log _{10} 18$
$=\log _{10} 8 \times 25 \times 9-\log _{10} 18 \ldots .\left[\log _a \mathrm{I}+\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\right.\left.\log _a \operatorname{Imn}\right]$
$=\log _{10} 1800-\log _{10} 18$
$=\log _{10}\left(\frac{1800}{18}\right)$
$=\log _{10} 100 \ldots .\left[\because \log _{10} 100=2\right]$
$=2$
$\log _{10} 8+\log _{10} 25+2 \log _{10} 3-\log _{10} 18$
$=\log _{10} 8+\log _{10} 25+\log _{10} 3^2-\log _{10} 18 \ldots .\left[n \log _a m=\log _a m^n\right]$
$=\log _{10} 8+\log _{10} 25+\log _{10} 9-\log _{10} 18$
$=\log _{10} 8 \times 25 \times 9-\log _{10} 18 \ldots .\left[\log _a \mathrm{I}+\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\right.\left.\log _a \operatorname{Imn}\right]$
$=\log _{10} 1800-\log _{10} 18$
$=\log _{10}\left(\frac{1800}{18}\right)$
$=\log _{10} 100 \ldots .\left[\because \log _{10} 100=2\right]$
$=2$
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