How do you condense $1/2log8v+log8n-2log4n-1/2log2j$ ?

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Answer 1 · 2016-06-19T20:53:07

$log(1/(n)sqrt((v)/j))$

By using log properties, you can write

$log(8v)^(1/2)+log(8n)-log(4n)^2-log(2j)^(1/2)$

and then, by grouping terms,

$log(sqrt(color(red)8v)/sqrt(color(red)2j))+log((color(red)8canceln)/(color(red)16n^cancel2))$

$=log(sqrt((color(red)4v)/j))+log(1/(2n))$

By using again log properties, you obtain

$log(1/(cancel2n)cancel2sqrt((v)/j))$

$log(1/(n)sqrt((v)/j))$

How do you condense 1/2log8v+log8n-2log4n-1/2log2j1/2log8v+log8n-2log4n-1/2log2j?