Question

Find the value of $\log 6+2\log 5+\log 4–\log 3–\log 2$.

Answer 1

$\log 6+2\log 5+\log 4–\log 3–\log 2$
$=(\log 6+\log 5^2+\log 4)–(\log 3+\log 2)$
$=\log (6\times 5^2\times 4)–\log 3\times 2$
$=\log \frac{6\times 5^2\times 4}{3\ast 2}$
$=\log (5^2\times 4)$
$=\log (25\times 4)=\log 100=2$
Hence, $\log 6+2\log 5+\log 4–\log 3–\log 2=2$