How do you condense $ln 3 - 2 ln (9 + 3)$ $ln 3 − 2 ln(9 + 3)$ ?

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Answer 1 · 2016-09-20T11:41:46

= $ln (\frac{3}{12^{2}}) = ln (\frac{1}{48})$ $ln(3/12^2) = ln(1/48)$

"If you are subtracting logs, then you are dividing the numbers"

$ln 3 - 2 ln (9 + 3)$ $ln 3 - 2ln(9+3)$

= $ln 3 - {ln (9 + 3)}^{2} \leftarrow$ $ln3 - ln(9+3)^2" "larr$ using the power law of logs

= $ln (\frac{3}{{(9 + 3)}^{2}})$ $ln(3/(9+3)^2)$

= $ln (\frac{3}{12^{2}}) \leftarrow$ $ln(3/12^2)" "larr$ 2 log terms have become one log term

We can now simplify further:

= $ln (\frac{3}{144})$ $ln(3/144)$

= $ln (\frac{1}{48})$ $ln(1/48)$

How do you condense ln 3 − 2 ln(9 + 3) ln3−2ln(9+3)ln 3 − 2 ln(9 + 3) ?