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

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Answer 1 · 2017-02-25T14:20:06

$4ln2-ln3-1=ln(16/(3e))$

$4ln2-ln3-1$

We know that: $alnb=lnb^a$ , therefore,

$=ln2^4-ln3-1$
$=ln16-ln3-1$

We know that: $lna-lnb=ln(a*1/b)$ , therefore,

$=ln(16*1/3)-1$
$=ln(16/3)-1$

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

$=ln(16/3)-lne$
$=ln(16/3*1/e)$
$=ln(16/(3e))$

Hence, $4ln2-ln3-1=ln(16/(3e))$

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