Question
Simplify the following:$2 \log 7+3 \log 5-\log \frac{49}{8}$
✓
Answer
$2 \log 7+3 \log 5-\log \frac{49}{8}$
$ =2 \log 7+3 \log 5-\log 49+\log 8$
$ =2 \log 7+3 \log 5-\log 7^2+\log 2^3 $
$ =2 \log 7+3 \log 5-2 \log 7+3 \log 2 $
$ =3 \log 5+3 \log 2 $
$=3(\log 5+\log 2)$
$ =3 \log (5 \times 2) $
$ =3 \log 10$
$=3 \times 1 $
$ =3 .$
$ =2 \log 7+3 \log 5-\log 49+\log 8$
$ =2 \log 7+3 \log 5-\log 7^2+\log 2^3 $
$ =2 \log 7+3 \log 5-2 \log 7+3 \log 2 $
$ =3 \log 5+3 \log 2 $
$=3(\log 5+\log 2)$
$ =3 \log (5 \times 2) $
$ =3 \log 10$
$=3 \times 1 $
$ =3 .$
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