How do you condense $ln 2 + 2 ln 3 - ln 18$ $ln2+2ln3-ln18$ ?

Question

How do you condense $ln 2 + 2 ln 3 - ln 18$ $ln2+2ln3-ln18$ ?

1 Answer

Impact of this question

14293 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-15T09:53:50

0

Explanation:

Using the 'laws of logarithms ' :

• logx + logy = logxy

• logx - logy = $log (\frac{x}{y})$ $log(x/y )$

• ${log x}^{n} = n log x$ $logx^n = n logx$

Applying these to this question :

$ln 2 + 2 ln 3 - ln 18 = ln 2 + {ln 3}^{2} - ln 18 = ln 2 + ln 9 - ln 18$ $ln2 + 2ln3 - ln18 = ln2 + ln3^2 -ln18 = ln2 + ln9 - ln18$

$= ln (\frac{2 \times 9}{18}) = ln (\frac{18}{18}) = ln 1 = 0$ $= ln((2xx9)/18) = ln(18/18) = ln1 =0$

Science

Math

Humanities

... and beyond

How do you condense $ln 2 + 2 ln 3 - ln 18$ $ln2+2ln3-ln18$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you condense ln2+2ln3−ln18ln2+2ln3-ln18?

1 Answer

Explanation:

Related questions

Impact of this question

How do you condense $ln 2 + 2 ln 3 - ln 18$ $ln2+2ln3-ln18$ ?