How do you condense #3 log_2 t -1/3 log_2 u+4 log_2 v#?

Question

1727 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-23T10:15:02

#log_2((t^3*v^4)/root(3)u)#

As bases are common and we can condense

#3log_2t-1/3log_2u+4log_2v#, using following identities

#alog_xb=log_xb^a#, #1/alog_xb=log_xb^(1/a)=log_xroot(a)b#

#log_xp+log_xq=log_xpq# and

#log_xp-log_xq=log_x(p/q)#.

Hence, #3log_2t-1/3log_2u+4log_2v#

= #log_2t^3-log_2root(3)u+log_2v^4#

= #log_2((t^3*v^4)/root(3)u)#