How do you condense $4 ln 2 - ln 3 - 1$ $4 ln 2 - ln 3 - 1$ ?

Question

3789 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-25T14:20:06

$4 ln 2 - ln 3 - 1 = ln (\frac{16}{3 e})$ $4ln2-ln3-1=ln(16/(3e))$

$4 ln 2 - ln 3 - 1$ $4ln2-ln3-1$

We know that: $a ln b = {ln b}^{a}$ $alnb=lnb^a$ , therefore,

$= {ln 2}^{4} - ln 3 - 1$ $=ln2^4-ln3-1$
$= ln 16 - ln 3 - 1$ $=ln16-ln3-1$

We know that: $ln a - ln b = ln (a \cdot \frac{1}{b})$ $lna-lnb=ln(a*1/b)$ , therefore,

$= ln (16 \cdot \frac{1}{3}) - 1$ $=ln(16*1/3)-1$
$= ln (\frac{16}{3}) - 1$ $=ln(16/3)-1$

We know that: $ln e = 1$ $lne=1$ , therefore,

$= ln (\frac{16}{3}) - ln e$ $=ln(16/3)-lne$
$= ln (\frac{16}{3} \cdot \frac{1}{e})$ $=ln(16/3*1/e)$
$= ln (\frac{16}{3 e})$ $=ln(16/(3e))$

Hence, $4 ln 2 - ln 3 - 1 = ln (\frac{16}{3 e})$ $4ln2-ln3-1=ln(16/(3e))$

How do you condense 4ln2−ln3−14 ln 2 - ln 3 - 1?