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If $f(x)=\log _{10}(1+x)$, then what is $4 \mathrm{f}(4)+5 f(1)-\log _{10} 2$ equal
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The correct answer is:
4
$\mathrm{f}(\mathrm{x})=\log _{10}(1+\mathrm{x})$
$4 \cdot \mathrm{f}(4)+5 \cdot \mathrm{f}(1)-\log _{10} 2$
$=4 \cdot \log _{10}(1+4)+5 \cdot \log _{10}(1+1)-\log _{10} 2$
$=4 \log _{10} 5+5 \log _{10} 2-\log _{10} 2$
$=4 \log _{10} 5+4 \log _{10} 2$
$=4\left(\log _{10} 5+\log _{10} 2\right)$
$=4\left(\log _{10} 10\right)=$
$4 \cdot \mathrm{f}(4)+5 \cdot \mathrm{f}(1)-\log _{10} 2$
$=4 \cdot \log _{10}(1+4)+5 \cdot \log _{10}(1+1)-\log _{10} 2$
$=4 \log _{10} 5+5 \log _{10} 2-\log _{10} 2$
$=4 \log _{10} 5+4 \log _{10} 2$
$=4\left(\log _{10} 5+\log _{10} 2\right)$
$=4\left(\log _{10} 10\right)=$
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