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

Question

1697 views around the world

You can reuse this answer
Creative Commons License

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))#