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43 votes
43 votes
Log14/3 +log11/5-log22/15=log

Log14/3 +log11/5-log22/15=log-example-1
User Albertodebortoli
by
2.6k points

2 Answers

25 votes
25 votes

Answer:

log 7

Explanation:

Adding the first values :

Applied Rule : log a + log b = log(ab)

⇒ log (14/3) + log (11/5)

⇒ log (14/3 × 11/5)

log (154/15)

Subtracting the second values :

Applied Rule : log a - log b = log(a/b)

⇒ log (154/15 ÷ 22/15)

⇒ log (154 ÷ 22)

log 7

User Hans Dash
by
2.6k points
23 votes
23 votes

Answer:


log7

Explanation:

when adding logs, apply the log rule:
\log _a\left(x\right)+\log _a\left(y\right)=\log _a\left(xy\right)


\log\left((14)/(3)\right)+\log\left((11)/(5)\right)=\log\left((14)/(3)\cdot (11)/(5)\right)

when subtracting logs, apply the log rule:
\log _a\left(x\right)\:-\:\log _a\left(y\right)=\log _a\left((x)/(y)\right)


\log\left((14)/(3)\cdot (11)/(5)\right)-\log\left((22)/(15)\right)=\log\left(((14)/(3)\cdot (11)/(5))/((22)/(15))\right)\\\\=\log\left(7\right)

User XtremeBaumer
by
2.8k points